British author Lady Gwendolen Gascoyne-Cecil encrypted a passage in her diary. This cryptogram has never been solved.

Enrypted diaries are a fascinating topic, but they rarely make major crypto mysteries. This is because the ciphers used by diary authors are usually simple substitutions (MASCs), which are easy to solve for a skilled codebreaker. Applying a more secure encryption system is normally not an option, as complicated ciphers are too laborious for writing longer journal entries. For this reason, my encrypted book list doesn’t contain any unsolved encrypted diaries.

Nevertheless, there is at least one nice crypto mystery connected to a diary. This journal was written by Lady Gwendolen Gascoyne-Cecil (1860-1945), the daughter of British Prime Minister Robert of Salisbury (1830-1903). Besides being a member of a prominent family, the Lady’s claim to fame was a successful book she wrote about her father.

Source: Wikimedia Commons

A few years ago, blog reader Ralf Bülow kindly pointed out to me that Lady Gewndolen had left behind an interesting cryptogram in her (otherwise unencrypted) diary. The source Bülow mentioned was the German book Wilhelm II. written by historian John C. G. Röhl (check here for the corresponding passage on Google Books).


A diary entry with numbers

Lady Gwendolen wrote the diary entry in question in the year 1888, which went down in German history as the Year of Three Emperors. In her journal, she mentions a person she describes as “465113, 49359”. According to the book by Röhl, a passage with eight more number groups follow. It is not known what these encrypted passages mean. I blogged about this story in 2015 (in German), but my readers couldn’t solve the mystery.

In a comment, blog reader Richard SantaColoma remarked that the diary excerpt reproduced in Röhl’s book was incomplete. Richard found a longer (complete?) version of this text in the book Salisbury: Victorian Titan by Andrew Roberts.

Now, four years later, I would like to address Lady Gwendolen’s diary again. This time, I quote the journal entry as reproduced in Roberts’ book (hoping that this is the correct version). Here it is:

S[alisbury]’s political anxiety is enormously increased by a conversation he had with 465113, 49359 — just 54461 415211 A & 55154 — is so 39751 30939 the 62818 562412 has been 60313 that 525210 the first 62041 562412 will 47651 497316 her 52062 will be 59941 60633 and yet 546410 477316. S[alisbury] could hardly believe his ears and was still more horrified when he gathered that 535611 58955. During 443513, 41463, always 39869 39869 31763 43447 and family and 48811 seems 50458 29911. From what S[alisbury] knows 457316 434447 336318 43168 thinks 43447. 33145 497316 anything, though this 50741 34942. He does not think that 562412 63131 58453 if she 60633. Doubtful how far 43447 45147 over 44148 40012 even now, and if he should 35234 S thinks 48355 36946 incapable 497316 424219 & 47651 — 539620.

As a conclusion, Roberts writes:

Any clue as to which code was being employed has sadly not survived. Despite the best efforts of the Foreign Office Librarian, of the decrypters, GCHQ, and many kind readers of The Times’ literary pages, including Bletchley Park alumni, it has proved impossible to decipher it, owing to its shortness. Something of the sense of crisis can be deduced from the vocabulary which intersperses the numbers, however.

When Lady Gwendolen wrote this diary entry, Wilhelm I (the first of the three emperors ruling that year) had already died. His son and successor Friedrich III deceased a few months later of throat cancer. Was it the news about the serious disease of the new emperor that shocked Salisbury? This sounds like a plausible hypothesis. However, when Friedrich III took office, his suffering was no longer a secret – his illness was already so advanced that he could not speak any more.


Is the encryption solvable?

My first guess was that Lady Gwendolen had used a code for encrypting her diary entry and that the numbers represent codewords. Codebooks were a common means of encryption at that time. In addition, each number group consists of five or six digits, which is consistent with a code, too. The following scan shows a codebook page (it’s certainly not the one used by Lady Gwendolen; it contains number groups, not letter groups):

Source: Internet Archive

If the code hypothesis is correct, each number in the diary passage stands for a whole word or an expression of several words. Solving the cryptogram will probably only be possible if somebody finds the codebook.

However, after my first post adressing the Gwendolen cryptogram, blog reader Tobias Schrödel wrote that the numbers the Lady used are too high for a typical code of the late 19th century. Codebooks of this era had at most 30,000 entries, so the codegroups usually consisted of low five-digit numbers.

If Lady Gwendolen didn’t use a code, which other encryption method might she have applied? Was it a simple substitution (MASC)? In don’t think that back in 1888 a person without a military background or a special interest in ciphers used a much more complicated system. So, it seems possible that the encryption can be broken. On the other hand, there is not much ciphertext to study, which makes it difficult to attack even a simple encryption method.

If a reader can find out more, please let me know?

Further reading:

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Kommentare (24)

  1. #1 Magnus
    1. August 2019

    The group 39869 is occurs twice, one after another. Is that a typo?

    43447 occurs three times in the entry, and there is a similar group, 434447, occuring once. Is that so in the original?

    Could this somehow be a book cipher perhaps?

  2. #2 Tony Patti
    2. August 2019

    It strikes me as quite non-random that the first digit only ranges from 2 through 6 (inclusive), and is unevenly distributed:

    first digit 2 occurs once
    first digit 3 occurs 10 times
    first digit 4 occurs 23 times (the most)
    first digit 5 occurs 15 times (second most)
    first digit 6 occurs 6 times

    As to Magnus‘ comment (above), perhaps the first digit could be chapter of a book?

    Or alternatively, it is the first two digits ranging from 29 through 63 (inclusive) which is important, which would somewhat explain why the only occurrence of first digit two (29911) is so close to first digit “3”.

    The non-randomness of the first digit can be contrasted with the last digit which fully ranges from zero through nine.

  3. #3 Tony Patti
    2. August 2019

    I also note the non-randomness of the next-to-last digit which ranges only from 1 (most common value) to 6:

    next-to-last digit 1 occurs 21 times
    next-to-last digit 2 occurs only once
    next-to-last digit 3 occurs only 5 times
    next-to-last digit 4 occurs 12 times
    next-to-last digit 5 occurs 9 times
    next-to-last digit 6 occurs 7 times

  4. #4 Thomas
    2. August 2019

    Pretty sure it´s a book code, as Magnus wrote:

    The first three digits stand for the page, the 4th for the xth line an the 5th and 6th (in 6-digit-numbers) for the yth word in this line. What catches the eye ist that the 5th digit in each 6-digit-number is “1”, there is only one exception in the last number, there it is “2”, followed by “0”, so that there are max. 20 words in one line of the book. Only “434447” doesn´t match this scheme. But – since there are three “43447”, I think his is a typo with a “4” to much. Judging from the range of the first three digits of each number, the the codes were taken from a chapter in the book running approx. from page 308 – 629. This must be a chapter containing the names of her contemporaries mentioned in the diary (obviously a name: 465113), hence a chapter on current politics.

  5. #5 Thomas
    2. August 2019

    Maybe the book used by her could be found here:

  6. #6 Magnus
    2. August 2019

    How many pages is there in the diary, maybe it is the key? As Thomas says, the source must have names of her contemporaries. What better than the diary itself?

    Also, the group 40012 further indicates that the three first digits belong together since it would be unlikely for something to start with one or more zeroes.

  7. #7 Thomas
    2. August 2019

    Good idea! How many pages? No idea: The ‘Lady Gwendolen Cecil Papers’ and her diary ( are kept by Hatfield House (see link 5#) and aren’t available online. So only anyone who stays in or near Hertfordshire could have a look at it or look for other possible sources in the family library. But I’d assume that any of the experts who had worked on the encrypted passage (Salisbury’s biographer Roberts: “Despite the best efforts of the Foreign Office Librarian, of the decrypters, GCHQ, and many kind readers of The Times’ literary pages, including Bletchley Park alumni, it has proved impossible to decipher it, owing to its shortness.”) has already taken a book code into account and searched for a source in Hatfield House.

  8. #8 Klaus Schmeh
    2. August 2019

    Christof Rieber via Facebook:
    All cipher sections appear to have either five or six digits, appearing quite unusual for cleartext. Thus, no substitution, imo. All digits should be put together, as a separate message, imo.

    Cipher then reads as:

    46511 34935 95446 14152 11551 54397 51309 39628 18562 41260 31352 52106 20415 62412 47651 49731 65206 25994 16063 35464 10477 31653 56115 89554 43513 41463 39869 39869 31763 43447 48811 50458 29911 45731 64344 47336 31843 16843 44733 14549 73165 07413 49425 62412 63131 58453 60633 43447 45147 44148 40012 35234 48355 36946 49731 64242 19476 51539 620

    Cipher length 293.

    Very clearly, fake language text has been put around these numbers, to hide the encryption (as one complete message independently from the text itself); which now may or may not be solved.

    Plenty of double homophones (24, more than 8%..), with one section even repeating (39869 39869, transcription error possible).

    Also, 73164 then 73165 occurring..chance of ~1:75,000…


  9. #9 Klaus Schmeh
    5. August 2019

    Michael Schroeder via Facebook:
    Pollux cipher?

  10. #10 Klaus Schmeh
    5. August 2019

    Christof Rieber via Facebook:
    Great idea, same time and would explain the double letters..

  11. #11 Klaus Schmeh
    5. August 2019

    Christof Rieber via Facebook:
    Numbers 4, 1, 3, 5, 6 cover approximately 70% (!) of the cipher..

  12. #12 Magnus
    8. August 2019

    I have some more information on the diary.
    The diary is in one volume (from 1888) and it has 174 pages. The pages are numbered.

    The encrypted section starts on page 71.

    So that rules our that the three first digits of the code groups would refer to pages of the diary itself, unfortunately.

    It is still feasible for the two first digits to refer to a page of the diary, but then the question remains how the following three to four digits should be interpreted…

  13. #13 Magnus
    8. August 2019

    I can also say that the line “From what S[alisbury] knows 457316 434447 336318 43168 thinks 43447.” is not correct.

    In the diary it says:
    “From what S knows 457316 43447 336318 43168 thinks 43447” and then there is a smudge that has been interpreted as a period, but I would say that it isn’t. So in the diary there is no code 434447: it is 43447.

  14. #14 Klaus Schmeh
    9. August 2019

    Christof Rieber via Facebook:
    Meanwhile quite sure that we deal with a STRADDLING CHECKERBOARD cipher.

  15. #15 Thomas
    11. August 2019


    That’s interesting news. Now we can be sure it’s a book code. The last digit of the five digit groups (1 – 9 ) and the second to last and the last digit of the six digit groups (11 – 20) indicate which word in a line has to be taken. It’s no coincidence that the second to last digit in each six digit group is “1”, this rules out substitution and transposition ciphers. Moreover, as you’ve found out and as I’ve assumed, the only exception “434447” is a mistake. What do the preceding digits indicate? The third to last should be the line containing the word that has to be chosen. Whether the page numbers consist of three digits (which would hint at a chapter running from the 300s to the 600s) or of only two digits and the first digit stands for the volume – who knows? Did you find out if she had already written a diary in previous years and every year has a (numbered) volume? E.g. 1884 – 1887 = volumes 3 – 6.

  16. #16 Magnus
    12. August 2019

    Yes, I’m quite sure it is a book code, and I’m also quite confident that the numbers should be interpreted (aaa)(b)(cc) or (aaa)(b)(c).

    With “b” ranging from 1-6 and “c” ranging from 1-20 I would rather let “c” denote the line number and “b” the word on that number. Coincidentally this is approximately the format of the diary itself: there are about 6-7 words per line and about 20-22 lines per page (for the few pages that I have seen photos of).

    Tha “a” numbers are fairly uniformly distributed between 299 and 631. The number 400 is included here which kind of suggests that the three digits should be seen as one number. I don’t think she numbered her first diary as “0”, but that’s just my guess of course.

    The entries in the diary have a date which is rather unsurprising. But she has also numbered the pages. Why would one write page numbers in a diary – the dated entries should be good enough? Using the page numbers in a code is perhaps a good reason…

    The first page of the diary is dated February 10. This suggests that more than one volume or more diaries did once exist. Hatfield house archives only have once diary, and that is this one from 1888. I can’t say if she numbered her diaries or not.

    Since the diary only have 172 pages, and the encoded message appears around page 72, I doubt that the diary itself was used as the key. It would undoubtedly have been convenient for the author to use the same book as a key – it will always be in the same location as the coded message. I guess we might need to look for another book that has more pages, unless one of the numbers in “a” is a null, or something like that.

    What other books with names of contemporary persons would have been easily accessible at the time? Some sort of year book? Almanac?

  17. #17 Klaus Schmeh
    31. August 2019

    Christof Rieber via Facebook:


    465113493595446141521155154397513093962818562412603135252106204156241247651497316520625994160633546410477316535611589554435134146339869398693176343447488115045829911457316434447336318431684344733145497316507413494256241263131584536063343447451474414840012352344835536946 49731642421947651539620

    With 293 figures, plenty of classic encryption methods can be ruled out: 293 is a prime number, so there won’t work any bigram analysis or patterns with e.g. 5 or 7 columns…same with straddling checkerboard – won’t work.

    Back to Morse code?

  18. #18 Juha
    2. September 2019

    This way comes first to my mind. For example:
    dice i

    But, perhaps this isn`t right way to solve

  19. #19 Juha
    9. September 2019

    Esperanto fits fairly well, how to crack this out.
    For example (using Google):

    465113 = (d) feaac = bean
    49359 = dic ei= I say to them

    ..and so on..

  20. #20 Juha
    14. September 2019

    Try this way:

    o=0, a=1, b=2, c=3, d=4, e=5, f=6, g=7, h=8, i =9…u=21…x=24

    all numbers do not fit
    -41463 dadfc (take just) dad
    -3986939 8 69 3176343447 cihfici fi= be scientific
    -48811 dhhaa (laugh)
    -50458 29911 odeh biik / (odeh bi)
    -316 caf= (coffee) or a person
    -525210 ebebao = (ebony, a tree ) or a person
    -539620 that it´s good

    -Esperanto was published 1887 (a book) and the diary is written 1888.
    -The grammar is not the main issue in diaries usually.

  21. #21 Klaus Schmeh
    1. Oktober 2019

    Christof Rieber via FaceBook:
    Another update…in her encrypted text, Lady Gwendolen used only two types of digits: 5-digit and 6-digit numbers (above transcription shows all numbers combined to each other).

    If we translate that sequences into short (5-digits) and long (6-digits), we might get the following Morse code out of Gwendolen’s notes:


    The problem with Morse code without spaces is its interpretation of letters.

    I wrote a program in Python with Aho-Corasick algorithm to do both, consume the morse code with all possible combinations as well as (simultaneously) searching words from an English (root) dictionary.

    Not yet too many promising results, but when searching for 5-letter plus words in the first, second, third part of the cipher, the following table can be shown:

    Still not sure if it is a Morse code at all, however.

    Example (partial)

  22. #22 Christof Rieber
    5. Oktober 2019

    Codebook cipher:

    “Salisbury’s political anxiety is enormously increased by a conversation he had with Bismarck, Friedrich – just crowned as a regent – is so sick that the Emperor’s lady has been announcing that Augusta, the first married lady, will abdicate of. Her grandson will be the successor and yet got empowered. Salisbury could hardly believe his ears and was still more horrified when he gathered that Wilhelm I died. During conversation, Robert, always spoke. Spoke about Wilhelm II and family and he seems very upset. From what Salisbury knows of Wilhelm II, sister Victoria thinks Wilhelm II, unaware of anything, though this not true. He does not think that lady will reign if she successor. Doubtful how far Wilhelm II rules over her decisions even now, and if he should decline. Salisbury thinks she is incapable of politics and leadership – Gwendolen.”

  23. #23 Thomas
    6. Oktober 2019

    @Christoph Rieber

    How have you figured this out? Did you find the codebook?

  24. #24 Christof Rieber
    Vienna / Budapest
    7. Oktober 2019

    I still even can’t be sure it is 100% correct, eg have problems with the word ‘successor’ used as a verb and noun at the same time. Also, the names could be different, too. Besides that, after starting with ‘of’ (incapable xxxxx), the rest got together in a way that the numbers that repeat actually had not only fitted well but came the overall context of the message, too. Thus, it is just a suggestion how the codebook might have been like.