Heute stelle ich einen neuen Preis für erfolgreiche Codeknacker vor: den Friedman-Ring. Wer ein von Armin Krauß entwickeltes Kryptogramm löst, kann ihn gewinnen.

English version (translated with DeepL)

Viele Leser werden sich erinnern: Am 27. April 2021 bloggte ich über ein Krypto-Rätsel, das mir Buchautor Frank Schwellinger zugesandt hatte.

Quelle/Source: Schwellinger

Nach mehreren Wochen und insgesamt 102 Leser-Kommentaren war die Lösung gefunden. Anna Salpingidis und Christoph Tenzer konnten sich den von Frank Schwellinger gestifteten Preis von 100 Euro unter sich aufteilen.

Quelle/Source: Salpingidis/Tenzer

Anna Salpingidis kam anschließend auf die Idee, selbst ein Krypto-Rätsel zu entwickeln. Dieses “Annagramm” veröffentlichte ich am 8. Juli 2021:

Quelle/Source: Salpingidis

Auch dieses Mal blieben meine Leser nicht untätig. Armin Krauß, einer der besten Codeknacker, die ich kenne, war der erste, der die korrekte Lösung präsentierte.

Quelle/Source: Krauß

Kommentare, die Armin möglicherweise geholfen haben, kamen von Thomas Bosbach, Karo und Frank Schwellinger.

 

Der Friedman-Ring

Nun hatte ich eine Idee. Nachdem Armin Krauß das zweite Rätsel aus dieser Serie gelöst hatte, wäre es doch logisch, wenn er das dritte selbst erstellen würde.

Mich erinnerte das an den Iffland-Ring. Dabei handelt es sich um einen Preis für Schauspieler, der auf den Schauspieler, Dramatiker und Theaterdirektor August Wilhelm Iffland (1759–1814) zurückgeht. Der Iffland-Ring wird jeweils testamentarisch vom Vorbesitzer an den seiner Meinung nach besten lebenden männlichen Schauspieler des deutschsprachigen Theaters verliehen. Der neue Träger behält den Preis bis zu seinem Tod und vererbt ihn seinerseits an den von ihm gewählten Nachfolger.

Der aktuelle Träger des Iffland-Rings ist seit 2019 Jens Harzer, den ich zugegebenermaßen nur aus diesem Zusammenhang kenne. Von 1996 bis zu seinem Tod war Bruno Ganz der Inhaber des Preises. Dieser spielte unter anderem in den Filmen “Der Untergang” und “Der Baader Meinhof Komplex” mit. Auch Ganz’ Iffland-Ring-Vorgänger Josef Meinrad war mir aus verschiedenen Fernsehserien bekannt.

Seit 1978 gibt es mit dem Alma-Seidler-Ring einen vergleichbaren Preis für Schauspielerinnen.

Analog zum Iffland- und Alma-Seidler-Ring rufe ich hiermit den “Friedman-Ring” ins Leben. Er ist nach dem Ehepaar Elizebeth und William Friedman benannt, die beide als Ausnahme-Codeknacker in die Geschichte eingingen. Natürlich wird dieser Preis gleichermaßen an Männer und Frauen verliehen.

Als ehemalige Träger des Friedman-Rings lege ich rückwirkend Frank Schwellinger, Anna Salpingidis und Christoph Tenzer fest. Der aktuelle Träger ist Armin Krauß.

Der jeweilige Empfänger des Friedman-Rings verpflichtet sich, ein Krypto-Rätsel zu entwickeln und mir zur Verfügung zu stellen. Wer dieses Rätsel als erstes löst, ist der neue Träger.

Noch gibt es den Friedman-Ring nur virtuell. Möglicherweise werde ich aber einen realen Ring basteln oder basteln lassen, der dann von einem Träger an den nächsten weitergereicht werden kann. Vielleicht hat ein Leser eine Idee, wie man eine solche Trophäe herstellen könnte.

Bei den bisherigen Challenges gab es Geld- oder Sachpreise zu gewinnen. Allerdings möchte ich nicht jedem Löser zumuten, einen Preis zu stiften. Daher gibt es für Friedman-Ring-Träger vorerst nur Ruhm und Ehre zu gewinnen.

 

Armin-Krauß’ Friedman-Ring-Rätsel

Nach all den Vorreden wird es nun Zeit für das von Armin Krauß erstellte Rätsel. Hier ist es:

Quelle/Source: Krauß

Armin hat mir mitgeteilt, dass der Schwierigkeitsgrad seiner meiner Meinung nach vergleichbar mit den vorherigen Challenges ist, also nicht zu leicht, aber auch nicht zu schwer. Wenn die Lösung zu lange auf sich warten lässt, wird Armin Hinweise geben.

Zu den Nachteilen bei einer solchen Challenge gehört, dass jeder Leser-Hinweis, der als Kommentar veröffentlicht wird, einem anderen Codeknacker zum Gewinn verhelfen kann. Ich bitte mein Leser daher, nicht ganz so egoistisch zu sein. Wenn im Forum über Lösungswege diskutiert wird, macht das die Sache für alle spannender. Natürlich werde ich jeden Kommentar, der zur Lösung beigetragen hat, entsprechend würdigen.

Davon abgesehen kann ich nur sagen: Das Rennen ist eröffnet.


Further reading: A challenge cipher published by the FBI

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Kommentare (116)

  1. #1 Richard SantaColoma
    https://proto57.wordpress.com/
    29. Juli 2021

    As for a ring, let me sketch up a design and email it to you.

  2. #2 Magnus Ekhall
    Borensberg
    29. Juli 2021

    Here is my transcribed ciphertext, starting from the small dot:

    XMFYNGKMXTWNHNHJFVLTFZIHUWWAUXKYFKMXF
    BACXVQLHKFJOOSQDAJVEQYYFXFRTPALKISJGK
    NUSMHKVGCBNKXOTIKLCZEVEUYBWIFXPJBZB

  3. #3 Klaus Schmeh
    29. Juli 2021

    @Magnus: Thanks, this will make things easier.

  4. #4 TWO
    29. Juli 2021

    D and R? DR who?

  5. #5 TONY PATTI
    USA
    30. Juli 2021

    Hello Klaus, I would suggest that the physical ring should embody some cryptographic property, something like a “decoder ring”, and perhaps 3D-Printed, I found a few suggestions via Google search:
    https://www.thingiverse.com/thing:17582
    https://www.thingiverse.com/thing:14891
    https://www.thingiverse.com/thing:1706210
    https://www.prusaprinters.org/prints/4725-caesar-cipher-decoder-ring-braille-version
    Enjoy,
    Tony

  6. #6 Michael BGNC
    Lake Constance
    30. Juli 2021

    The moon. Hmmm… Tycho Brahe, Kepler, Copernicus or Mare Tranquillitatis ?!?

  7. #7 Michael BGNC
    Lake Constance
    30. Juli 2021

    Very low IoC. All letters of the alphabet are present. No abnormalities in the hex dump of the image file.

  8. #8 Ranthoron
    30. Juli 2021

    Some thoughts about the ring:
    Does it have to be a finger-ring, or would an arm-ring also work?
    And as overrated as Dan Brown is, he described some ideas that could work for such a ring (e.g. the sub rosa thing with the chest).

  9. #9 Michael BGNC
    Lake Constance
    31. Juli 2021

    I hope that Armin’s hints do not consist of pieces of plain text. That this doesn´t lead to anything, without any clue regarding the method, you can see in the front of the cafeteria of the George Bush Center for Intelligence. Ok, maybe I’m a little impatient, but when Armin thinks it’s “of medium difficulty”, I guess its impossible … LOL

  10. #10 Michael BGNC
    Lake Constance
    31. Juli 2021

    Due to the low IoC, it is of course a substitution cipher …

  11. #11 Gerd
    31. Juli 2021

    Without any knowlegde about the method, it seems impossible. But it was designed as a puzzle, so maybe there are hints we didn’t recognize yet. The moon, and the typeface of the writing, which is “NASA Font”.

  12. #12 Michael BGNC
    Lake Constance
    31. Juli 2021

    I also immediately recognized the NASA font, and the crater Tycho due to its ray system and the image editing.

  13. #13 Michael BGNC
    Lake Constance
    31. Juli 2021

    Ok. There is a big misaligned letter K ! This separates “KISJGKNUSMHKVGCBNKXOTIKLCZEVEUYBWIFXPJBZB” at the end of the text. Maybe part of the key…

  14. #14 Martin
    31. Juli 2021

    Ich bin kein Krypto-Experte, aber folgende Idee:
    Das Kryptogramm ist ein Ring. Vielleicht braucht es mehrere Runden, es zu entziffern! Z.B. jeden x-ten Buchstaben mehrmals im Kreis, beginnend nach dem Punkt.
    Dafür spricht auch, daß es 109 Buchstaben sind. Dies ist eine Primzahl. D.h. egal wie groß x als Schrittlänge gewählt wurde, werden alle Buchstaben einmal drankommen!

  15. #15 Gerd
    31. Juli 2021

    >Dafür spricht auch, daß es 109 Buchstaben sind.
    interesting hint, the number of letters is prime! Going around the circle and taking every xth letter is a transposition. We know already that there must be also a substitution. Z-340 was also a combination of both, and was solvable.

  16. #16 Armin Krauß
    1. August 2021

    As Gerd said in #11, it would be very hard wihout some hints, and I have embedded some in the image. You have found one, but not all.

  17. #17 Michael BGNC
    Lake Constance
    2. August 2021

    It is difficult to distinguish between the compression artifacts of the PNG format and real anomalies. So far I think I have found 3 abnormalities:

    1. B.
    2. The raised K
    3. The S under crater TYCHO

    What is very difficult, is to distinguish whether the spacing of the letters is wrong in certain places, eg. the spacing between IFXP JBZB.

  18. #18 Armin Krauß
    2. August 2021

    @Michael:
    1. Yes
    2. Yes
    3. No
    Also the letter spacing is of no importance.

  19. #19 Michael BGNC
    Lake Constance
    2. August 2021

    The moon has a rotational offset against the original image. Tycho is in the middle but should be further to the right …

  20. #20 Armin Krauß
    2. August 2021

    @Michael: Also not important. Work with the 2 hints you found.

  21. #21 Michael BGNC
    Lake Constance
    2. August 2021

    Ok! Thanks …

  22. #22 YeS
    2. August 2021

    Is the plaintext in English?

  23. #23 Karo
    Otterfing
    2. August 2021

    @Michael
    There is a V in the right corner, which is the other way round
    Between B and C
    Maybe another hint?

  24. #24 hias
    2. August 2021

    Das K sieht aus wie die Kreiszahl Pi und das B. wie eine 3.
    Ein Hinweis auf 3.1415… ?

  25. #25 Karo
    Otterfing
    2. August 2021

    There are 2 more. In the middle down Tycho and left corner

  26. #26 YeS
    2. August 2021

    @Karo
    Inverted V’s are A’s

  27. #27 Michael BGNC
    Lake Constance
    2. August 2021

    @Karo: In NASA Font this is the A !

  28. #28 Narga
    2. August 2021

    Taking Martin’s idea from #14 a step further:
    If you start at the raised K, take every n-th letter (clockwise) and want to end up at the “B.” at the top, it follows that n has to be 69. Which seems to fit the NASA+MOON theme. Could be a funny coincidence, though…

  29. #29 Gerd
    3. August 2021

    The raised K is the 69th letter, so this might be a clue. The circular transposed text from every 69th is:

    KUUQVXFKUXPTUZANCQFJYWFFIOTHAZLAEEF
    KKGNMXRHCDHGVMSKBYTTOXMBBAWVVJNHN
    KKFFILQNVXXIXYYLOJMSFJPWEJHBLYGFIXZKS
    WKCB

    and now?

  30. #30 Michael BGNC
    Lake Constance
    3. August 2021

    The lack of information about the method used reminds me of something, but I just can’t figure it out. And as always, when the possibilities increase exponentially to infinity, after a certain time I lose interest in dealing with them, but luckily it’s not that far yet. Fortunately, Armin doesn’t charge $ 50 for a useless back and forth.

    The extremely low IOC indicates a very good (long) passphrase. We have a subdivision of the given text. So it can be assumed that Armin provided us with the key or at least parts of it. I have no clue about the method used, but for the sake of simplicity, and because only capital letters are given, I will assume mod 26 for now.That will change the more information becomes known. I assume the method is solvable with pen and paper. For now I also exclude fancy things like ASCII binary XOR or CCITT2. There is no evidence of something about this

  31. #31 Armin Krauß
    3. August 2021

    @Michael: Well, I’m a bit hesitant in giving a more obvious hint because all the pieces of the puzzle have been identified by now. Now it’s only a question of
    putting them in the right order. Be assured that I haven’t used anything esoteric, only standard methods.

    I will say that much: there is one comment that has made a very good observation, especially in connection with your last post, but it seems it hasn’t been recognized yet.

  32. #32 YeS
    3. August 2021

    Based on Armin’s and hias comments my guess is that it’s some kind of running key cipher with digits of Pi serving as a key. I suppose the key should start from a certain digit of Pi. Also there’s no guarantee that ciphertext begins after the small dot – could begin anywhere.

  33. #33 Thomas
    3. August 2021

    @MichaelBGNC

    “Due to the low IoC, it is of course a substitution cipher”: Not necessarily: If the plaintext is much shorter than 109 letters, the ciphertext might comprise the plaintext (maybe transposed according to a certain scheme) and a certain number of nulls, yielding a low IoC.

  34. #34 Michael BGNC
    Lake Constance
    3. August 2021

    @Thomas: It’s great that the discussion thread generates so many ideas. This means that many different approaches are processed at the same time.

  35. #35 Gerd
    3. August 2021

    Hias’ Pi hint might be valuable. However I tried several approaches using CADAEIBFECEI or similar as a key and could not find any cleartext.

  36. #36 Michael BGNC
    Lake Constance
    3. August 2021

    My guess is that

    XMFY… is the message
    KISJ… is (part of) the long key, that has to be combined in some way with PI.

  37. #37 Michael BGNC
    Lake Constance
    3. August 2021

    @Gerd:

    Or DBEBFJCGFDFIJHJDCDIEGCGEDDIDCHJFACIIEBJHB … 0 (A) does no shift und 3 (D) is the 4th position. 0,1,2,3,…

  38. #38 Michael BGNC
    Lake Constance
    3. August 2021

    And maybe Armin used an alphabet key that has something to do with the image. Something like:
    APOLBCDEFGHIJKMNQRSTUVWXYZ
    TYCHOBRAEDFGIJKLMNPQSUVWXZ
    FRIEDMANBCGHJKLOPQSTUVWXYZ

  39. #39 Armin Krauß
    4. August 2021

    @YeS #22
    Yes, English.

    @YeS #32, @Gerd #35, @Michael #37
    You’re on the right track, but consider that there is still another unused hint.

    @Michael #36
    Correct, the message starts with XMFY. But it is not separated in different parts. It’s all ciphertext.

    @Michael #38
    No.

  40. #40 Thomas
    4. August 2021

    “It’s all ciphertext”: So I concude the length of the plaintext is 109 (no nulls).

    “B.” looks like “3.”: This might be a hint that the digits od Pi (1415926…) should be assigned to the ciphertext letters XMFY…. I tried to shift the ciphertext letters back in the alphabet (mod. 26) according to the digits of Pi which yields WIETEE…. Alll in all many useful letters contrary to the ciphertext (e.g. 30 E, 14 N, 9 O ….). This could be the substitution part of the cipher. But yet I’ve no idea as for the transposition part.

  41. #41 Michael BGNC
    Lake Constance
    4. August 2021

    @Thomas

    WIETEE… index of coincidence is 0.073 this looks good. The letters L,P,Q,V,X are missing. Its not a monoalphabetic substitution.

    Assuming we have to do a transposition the missing L, V and P is strange for an English plaintext.

    Can someone confirm the missing letters.

  42. #42 Gerd
    4. August 2021

    Using #40 and 109 digits of pi, I get:

    WIETEEEHUOOEAEEHCNHNDTEEROTYNOFYDC
    ETESTBRMNCYHYENONIBAAOAHUUAODOTISKE
    ESDECHSSEYBNAATNHTGRDHHAYDOEORSOGE
    THJTTW

    Histogram:
    https://ibb.co/yqCDjXY

  43. #43 YeS
    4. August 2021

    @Armin Krauß
    Was it enciphered with Vigenere or something else?

  44. #44 Armin Krauß
    4. August 2021

    WIETEE… looks good. I believe the eagle will land soon 🙂 (no, that’s not a hint).

    @YeS: No Vigenere.

  45. #45 Michael BGNC
    Lake Constance
    4. August 2021

    @Gerd: Thanks, I have the same result…

  46. #46 YeS
    4. August 2021

    We didn’t use the Moon/Nasa/Space thing yet. Maybe the date of Moon landing plays into it somehow.

  47. #47 TWO
    4. August 2021

    I think it is about the exploration of deep cryptospace.

    To boldly decipher a message from an ancient alien civilization.

    Or is it a paring ticket because to Arnim for staying in the left lane of the Mily Way?

    Whjo knows the exact answer? Only AK!

  48. #48 TWO
    4. August 2021

    correction
    parking ticket Milky Way

  49. #49 Armin Krauß
    4. August 2021

    @Michael: Yes, the plaintext is missing L, P and V.

  50. #50 hias
    4. August 2021

    Ist das “D” in der unteren Mitte des Bildes nur unscharf gezeichnet oder ein Hinweis ?

  51. #51 Armin Krauß
    4. August 2021

    The “D” is not a hint. The only hidden hints were the “3.” on top position and the exposed “K” on a position further down.

  52. #52 Thomas
    4. August 2021

    My first guess as for the transposition part would be a circular transposition (like proposed by Gerd #29) of the substitution result “WIETEE…” (Gerd #42). But which stepwidth to chose? What would be the best stepwidth for reaching all 109 letters in the circle as fast as possible? (Reminds me of the Josephus problem).

  53. #53 hias
    4. August 2021

    109 Buchstaben rund um den Mond.
    Umfang des Mondes ca. 10900 km.
    Zufall?

  54. #54 Thomas
    4. August 2021

    Maybe, as Gerd #29 wrote, the distance between the start and the raised letter is a hint that the stepwidth is 69.

  55. #55 TWO
    4. August 2021

    K3 as in Kryptos K3?

    A hidden double transpose?

  56. #56 TWO
    4. August 2021

    Wietee
    90 -91

    EENEOTMEAUIDETDOGTEOCENERYBUTSSARD
    OTTOHTYCBHIHOESAGYSJEUEDTDTYNADEHNT
    ARHIHENOYSCOOOKCBHHOTWEAHRFENNAASE
    YNHEEW

  57. #57 Michael BGNC
    Lake Constance
    5. August 2021

    I tested several transposition techniques. Eg. Skip, Box-Rotation, Railfence. I also brute forced Row, Column, Row in all possible variations (Column, Row, Column etc.) up to key length 11. No success so far. I wonder if there is another clue. Maybe the circle in which the text is written…

  58. #58 Thomas
    5. August 2021

    @Michael BGNC
    Did you try to arrange the substituted letters in a circle, chose the W as the first letter and subsequently every 69th letter clockwise?

  59. #59 Michael BGNC
    Lake Constance
    5. August 2021

    @Thomas: I do it by hand, this drives me crazy, because it is very error prone. I often misscount. Maybe I write a little software. So far I have: WE NO H…

  60. #60 Gerd
    5. August 2021

    I tried several rotational transpositions, up to now with no success. You can do it in Excel or Libreoffice, using the “TEIL(” String-Funktion, and calculate the positions you want by modulo “REST(” operator.

  61. #61 Armin Krauß
    5. August 2021

    @Michael, Thomas: Yes, the circular shape of the ciphertext is important.

    Here is another hint: the first letter of the plaintext is W.

  62. #62 TWO
    5. August 2021

    World

  63. #63 YeS
    5. August 2021

    No L’s – no worLd

  64. #64 TWO
    5. August 2021

    I know but Word sounds so mundane

  65. #65 YeS
    5. August 2021

    If circular shape is important, maybe something with letters on opposing sides of diameter lines or something like that?

  66. #66 Michael BGNC
    Lake Constance
    5. August 2021

    #64 Bei uns im Allemanischen würde man sagen: “Hauptsach´ nen Dreck rausgschwätzt …”

  67. #67 TWO
    5. August 2021

    @61

    Is it bi-diectional?

  68. #68 Armin Krauß
    5. August 2021

    @67 What do you mean by that?

  69. #69 Michael BGNC
    Lake Constance
    5. August 2021

    Maybe it has to do with “Die Quadratur des Kreises” (Squaring the circle ) … because of Pi !

  70. #70 Armin Krauß
    5. August 2021

    @Michael: No, you are thinking too complicated. It is really just a simple operation that is widely used in cryptography.

  71. #71 Michael BGNC
    Lake Constance
    5. August 2021

    Thank you for that hint. I’ve been operating here with a set square and compass for half an hour… LOL

  72. #72 Gerd
    5. August 2021

    No more ideas. Maybe we should try this 😉

  73. #73 Michael BGNC
    Lake Constance
    5. August 2021

    @Gerd: BEST IDEA SINCE 6 DAYS ! Great …

    I´m out for today… Maybe I should have had a drink every time Armin said “No” … Prost.

  74. #74 TWO
    5. August 2021

    @68
    to change the direction of decryption

  75. #75 George Lasry
    5. August 2021

    Armin – would it be possible to list here all the hints you have been giving or confirming. It is very difficult to follow from this long thread. Thx

  76. #76 Armin Krauß
    5. August 2021

    @George

    Of course:

    The image contains two visual hints: the “3.” in top position and the exposed “K” further down. The “3.” pointed to the subtraction of the digits of pi from the ciphertext, leading to the intermediate ciphertext “WIETEE…”.

    The other hint has not yet been used.

    The circular shape of the ciphertext is of some importance.

    The first letter of the plaintext is “W”.

    The final step is nothing esoteric, but a simple operation that is widely used in cryptography.

  77. #77 Armin Krauß
    5. August 2021

    @Gerd #72

    Great find!
    But that was not my inspiration. 🙂

  78. #78 TWO
    5. August 2021

    #72
    Do you mean the K furher down clockwise?

    The B K sequence is repated anti clockwise.

  79. #79 Armin Krauß
    5. August 2021

    #78 Yes, clockwise.

  80. #80 TWO
    6. August 2021

    K could point to modulus 11 ?

  81. #81 Narga
    6. August 2021

    I think it is
    “We choose to go to the Moon in this decade and do the other things not because they are easy but because they are hard John F Kennedy” and I have two Ys (and the ending W) left as fillers.

  82. #82 Michael BGNC
    Lake Constance
    6. August 2021

    @Narga: Sounds good and makes sense. What method did you apply? … and of course no L, P and V.

  83. #83 Armin Krauß
    6. August 2021

    YES!!!
    Congratulations, Narga, you found the correct plaintext. Well done!

  84. #84 Michael BGNC
    Lake Constance
    6. August 2021

    And of course: CONGRATULATION!

  85. #85 Michael BGNC
    Lake Constance
    6. August 2021

    And also a big “THANK YOU” to Armin, for this great puzzle. Several days of pure fun … Thumbs up!

  86. #86 TWO
    6. August 2021

    Great puzzle.

    Can we have more of these?

  87. #87 Gerd
    6. August 2021

    Congratulations, Narga! and thank you to Armin, that was a lot of fun. Could someone please disclose the transposition method, I still don’t get it.

  88. #88 Thomas
    6. August 2021

    @Narga
    Congratulations, great job!

    A long shot: Could it be each 69th or maybe11th (K: 11th letter in the alphabet) letter anti-clockwise? But why two Ys as nulls?

    @Armin
    Awesome puzzle, this was great fun, thanks for that!

  89. #89 Narga
    6. August 2021

    Thanks everyone! And thank you Armin for the nice puzzle. Unfortunately, I don’t know the original transposition method and would also really like to know it. I tried standard columnar and route transposition schemes as well as going in circles (both ways, starting from all letters) with all step sizes, including a brute force search of step patterns (like stepping repeatedly 7,20,19,69 letters).

  90. #90 Thomas
    6. August 2021

    I’m wondering why three (including the W at the end) nulls are needed (like sometimes for instance in a complete columnar transposition ) if it is some sort of cipher based on a circle. What is special about the number 109 (compared to 106, 107 and 108)?

  91. #91 Armin Krauß
    6. August 2021

    @Narga
    I’m impressed that you found the plaintext without knowing the transposition. Could you explain how you did it?

  92. #92 Thomas
    6. August 2021

    Maybe comparing the letter frequencies to all Moon quotes?

  93. #93 Armin Krauß
    6. August 2021

    @Thomas @Gerd @Michael
    I will not disclose the transposition method right away, because I think it might be fun to try to reconstruct it now that you know the plaintext. However, if noone comes up with it in a few days, I’ll disclose it of course.

    #90
    109 is a prime, that is important. 107 is also a prime, but unsuitable in this case.

    #88
    The letter “K” is not important, it just happenes to be at position 69 of the ciphertext. So just use 69 and ignore 11.

  94. #94 Narga
    6. August 2021

    #91,#92

    Yes, I do anagramming on a sentence basis to solve transpositions. I collect a database of sentences from quotes and books where each sentence is stored in a way where all letters in that sentence are alphabetically sorted. Then one just needs to sort the ciphertext, too, and start comparing. It helps if the ciphertext is just one sentence 🙂

  95. #95 Armin Krauß
    6. August 2021

    Well, a good thing then that I had the shifting with Pi digits as a first step. Otherwise you would have solved it on the first day. 🙂

    Nochmals Glückwunsch, gut gemacht!

  96. #96 TWO
    #95
    6. August 2021

    Does it start with 108 69?

    WENOHNTYESTHTRYAEANEDY….

  97. #97 Armin Krauß
    6. August 2021

    Hm, no.
    The 108th letter of the intermediate ciphertext is T, so your solution would start with TE.
    But Narga has already found the correct solution.

  98. #98 TWO
    6. August 2021

    #97
    This is skip 109 letters and then count 69 letters back,

    Never mind. no doubt you willl reveal the method some day.

  99. #99 TWO
    7. August 2021

    As usual a typo.

    What I meant is skip the first 108 letters then count 69 letters backwards.

    ohn nedy is visible now. and the distance between those is 13.

    IMHO you are varing the steps of the transpose maybe those nice PI digits are used again.

    After each step peel off some letters and restart the process.

    Round 1 =109 letters.
    Round 2 = 107 letters.

    Probably this leads to nothing.

    But what hint is hidden by the K?

  100. #100 Armin Krauß
    7. August 2021

    You are right, the transposition does not use a constant step size, nor does it use Pi in any way. The varying step size is a simple consequence of the used transposition method. But that makes it sound more complicated than it really is.

    The letter K was raised above the other letters in the image to indicate that the position of this letter, namely position 69 in the ciphertext, is of some importance. The number 69 is the hint, not K.

    In fact, the only two numbers necessary to build the transposition are 69 and 109 (length of the ciphertext).

  101. #101 TWO
    7. August 2021

    Interesting but anyhing to do with math is not easy for me.

    There is only one (1) J and one F in the text and they are in the final result result J ohn F 3 letters apart .
    There are 4 C’s but only one of them is followed by a H.
    I wish I was not so lazy and actually count the distances and see what numbers show up.

  102. #102 TWO
    7. August 2021

    JFKM

  103. #103 YeS
    7. August 2021

    Armin, you need to reveal the transposition method in comment #109 🙂

  104. #104 Armin Krauß
    7. August 2021

    I will.
    So you have 4 tries left now. 🙂

  105. #105 YeS
    7. August 2021

    I’m out of ideas. All I can do is hasten the reveal with this comment.

  106. #106 hias
    8. August 2021

    Gedanken zur 69

    Tag der Mondlandung 21.7.1969
    Die 69. Primzahl ist 347
    Produkt der Primzahlen 3 und 23

    Aber auch mit diesen Zahlen erkenne ich kein Muster. Vielleicht noch ein Hinweis ob das W von “WE” das von Stelle 1 oder 109 ist?

  107. #107 Armin Krauß
    8. August 2021

    Die ersten 4 Buchstaben des Klartextes, “WECH”, stehen an den Positionen 1, 69, 74 und 92 des Zwischentextes “WIETEE…”.

  108. #108 hias
    8. August 2021

    1. Buchstabe = Position 1
    2. Buchstabe = Position 1×69 = 69
    3. Buchstabe = Position 69×69 = 4761 ( =Position 74)
    4. Buchstabe = Position 74×69 =5106 ( =Position 92)
    5 Buchstabe = Position 92×69 = 6348 ( =Position 26)
    6. Buchstabe = Position 26×69 =1794 ( =Position50)
    usw.
    usw.

  109. #109 Armin Krauß
    8. August 2021

    Yes, that is it!

    The transposition positions are determined by the modular exponents of 69!

    Thus the challenge is completely solved! Thank you to all participants, it was great fun for me too!

    Now I’m looking forward to Narga’s challenge. 🙂

  110. #110 Narga
    9. August 2021

    I wouldn’t ever have derived that transposition system, not even with the complete set of position numbers. So, great job on that, hias!

    Interesting that this sequence, like a fixed step size, does not lead to a loop, i.e. touches every number/position only once. Well done, Armin!

    I have already a few ideas on the stack for a new challenge, but it might still take a bit of work in order to match the quality of the previous challenges here.

  111. #111 Thomas
    9. August 2021

    @Armin

    I find the math behind your cipher interesting: How did you know that the length of the ciphertext has to be 109 so that three nulls are needed? Did you use an algorithm or an equation yielding the length?

  112. #112 Armin Krauß
    9. August 2021

    @Thomas

    I used a bit of modular arithmetic for this: If the powers of a number A generate all possible remainders modulo some prime number P, then A is called a primitive root of P. I wanted 69 to play a role in my challenge (for obvious reasons), so I chose A = 69. Then I had to find a prime number for which 69 is a primitive root. With numbers of such small magnitude, this can be done with brute force. I wrote a little program to do that, and the smallest prime bigger than my plaintext length was 109.

  113. #113 Klaus Schmeh
    10. August 2021

    CONGRATULATIONS TO NARGA! YOU’RE THE NEW BEARER OF THE FRIEDMAN RING!

    And thanks to all others who participated, especially Magnus, TWO, Michael, BGNC, Gerd, Martin, Karo, hias, YeS, Thomas, George Lasry, Rich, Ranthoron, and Tony! To my regret, there’s only one Friedman Ring that can be awarded.

  114. #114 TWO
    11. August 2021

    Why is 4761 equal to 74?;

  115. #115 Armin Krauß
    11. August 2021

    Because 74 is the remainder of 4761 divided by 109.

  116. #116 TWO
    11. August 2021

    That is brilliant!