Back in 1939, Alexander d’Agapeyeff wrote a tidy little book called “Codes and Ciphers” on cryptography history: though you can now buy it print-on-demand, cheap copies of the original book often come up on the various second-hand book aggregators (such as bookfinder.com), which is where I got my copy of the “Revised and reset” 1949 edition.

What is now generally understood is that d’Agapeyeff wasn’t really a cryptographer per se: he had previously written a similar book on cartography for the same publisher, and so thought to tackle cryptography.

On the very last page of the text (p.144), d’Agapeyeff dropped in a little cipher challenge, saying “Here is a cryptogram upon which the reader is invited to test his skill.

75628 28591 62916 48164 91748 58464 74748 28483 81638 18174
74826 26475 83828 49175 74658 37575 75936 36565 81638 17585
75756 46282 92857 46382 75748 38165 81848 56485 64858 56382
72628 36281 81728 16463 75828 16483 63828 58163 63630 47481
91918 46385 84656 48565 62946 26285 91859 17491 72756 46575
71658 36264 74818 28462 82649 18193 65626 48484 91838 57491
81657 27483 83858 28364 62726 26562 83759 27263 82827 27283
82858 47582 81837 28462 82837 58164 75748 58162 92000

This modest little cryptogram, now known as “the d’Agapeyeff Cipher“, has somehow remained unbroken for 70 years, and is often to be found alongside the Voynich Manuscript on lists of cipher enigmas.

The first thing to note is that adjacent columns are formed alternately from 67890 and 12345 characters respectively: which is a huge hint that what we are looking at is (in part, at least) a grid cipher, where each pair of numbers gives a position in a grid. If so, then we can throw away the “patristrocat” spaces between the blocks of numbers and rearrange them as pairs.

75 62 82 85 91 62 91 64 81 64 91 74 85 84 64 74 74 82 84 83 81 63 81 81 74
74 82 62 64 75 83 82 84 91 75 74 65 83 75 75 75 93 63 65 65 81 63 81 75 85
75 75 64 62 82 92 85 74 63 82 75 74 83 81 65 81 84 85 64 85 64 85 85 63 82
72 62 83 62 81 81 72 81 64 63 75 82 81 64 83 63 82 85 81 63 63 63 04 74 81
91 91 84 63 85 84 65 64 85 65 62 94 62 62 85 91 85 91 74 91 72 75 64 65 75
71 65 83 62 64 74 81 82 84 62 82 64 91 81 93 65 62 64 84 84 91 83 85 74 91
81 65 72 74 83 83 85 82 83 64 62 72 62 65 62 83 75 92 72 63 82 82 72 72 83
82 85 84 75 82 81 83 72 84 62 82 83 75 81 64 75 74 85 81 62 92 00 0[0]

The first hint that the order of these might have been scrambled (‘transposed’) comes from the two sets of tripled letters: 75 75 75 and 63 63 63. Five centuries ago, even Cicco Simonetta and his Milanese cipher clerks knew that tripled letters are very rare (the only one in Latin is “uvula“, ‘little egg’). The second hint that this is a transposition cipher is the total number of characters (apart from the “00” filler at the end): 14×14. If we discard the filler & rearrange the grid we get:-

75 62 82 85 91 62 91 64 81 64 91 74 85 84
64 74 74 82 84 83 81 63 81 81 74 74 82 62
64 75 83 82 84 91 75 74 65 83 75 75 75 93
63 65 65 81 63 81 75 85 75 75 64 62 82 92
85 74 63 82 75 74 83 81 65 81 84 85 64 85
64 85 85 63 82 72 62 83 62 81 81 72 81 64
63 75 82 81 64 83 63 82 85 81 63 63 63 04
74 81 91 91 84 63 85 84 65 64 85 65 62 94
62 62 85 91 85 91 74 91 72 75 64 65 75 71
65 83 62 64 74 81 82 84 62 82 64 91 81 93
65 62 64 84 84 91 83 85 74 91 81 65 72 74
83 83 85 82 83 64 62 72 62 65 62 83 75 92
72 63 82 82 72 72 83 82 85 84 75 82 81 83
72 84 62 82 83 75 81 64 75 74 85 81 62 92

This is very probably the starting point for the real cryptography (though the presence of tripled characters in the columns implies that it probably isn’t a simple “matrix-like” diagonal transposition. Essentially, it seems that we now have to solve a 14×14 transposition cipher and a 5×5 substitution cipher simultaneously, over a relatively small cryptogram – an immense number of combinations to explore.

However, we know that d’Agapeyeff wasn’t a full-on cryptographer, so we should really explore the psychological angle before going crazy with an 800-year-long brute-force search. For a start, if you lay out the frequencies for the 5×5 letter grid (with 12345 on top, 67890 on the left), a pattern immediately appears:-

** .1 .2 .3 .4 .5
6. _0 17 12 16 11
7. _1 _9 _0 14 17
8. 20 17 15 11 17
9. 12 _3 _2 _1 _0
0. _0 _0 _0 _1 _0

Here, the 61 (top-left) frequency is 0, the 73 frequency is 0, and the final nine frequencies are 3, 2, 1, 0; 0, 0, 0, 1, 0. I think this points to a 5×5 mapping generated by a keyphrase, such as “Alexander d’Agapeyeff is cool” (for example). To make a keyphrase into a 5×5 alphabet, turn all Js into Is (say), remove all duplicate letters (and so it becomes ALEXNDRGPYFISCO), and then pad to the end with any unused characters in the alphabet in sequence (BHKMQTUVWZ)

* 1 2 3 4 5
6 A L E X N
7 D R G P Y
8 F I S C O
9 B H K M Q
0 T U V W Z

For a long-ish (but language-like) keyphrase, rare characters would tend to get moved to the end of the block: which is what we appear to see in the frequency counts above, suggesting that the final few letters are (for example) W X Y Z or W X Z.

Yet 61 and 73 have frequency counts of zero, which points to their being really rare letters (like Q or Z). However, if you read the frequency counts as strings, 61 62 63 = 0 17 12, while 73 74 75 = 0 14 17: which perhaps points to the first letter of the keyphrase (i.e. 61) being a rare consonant, and the second pair being Q U followed by a vowel. Might 73 74 75 76 77 be QUIET or QUITE?

I don’t (of course) know: but I do strongly suspect that it might be possible for a cunning cryptographer to crack d’Agapeyeff’s keyphrase quite independently of his transposition cipher. It can’t be that hard, can it? ;-p

———-
Update: a follow-up post is here

### 15 thoughts on “The d’Agapeyeff Cipher…”

1. The Danish mathematician Julius Petersen wrote a series of popular articles on cryptography. He, too, included one that has gone unsolved till this day.

My algebra teacher is interested in the history of mathematics, and he gave me a copy of the articles once. I have no idea where I keep them anymore, but I’m sure he has them at hand, should you be interested.

2. Ross on May 25, 2011 at 3:46 am said:

The only phrase relevant to this cipher is the invitation itself. Here is a cryptogram upon which the reader is invited to test his skill.

Removing duplicates leaves 20 characters. Since the 0 row is likely nulls then there is a 5×4 reference matrix. I have tried direct plugging but no dice.

Heris acypt ogmun wdvkl.

Remaining letters (ignoring j) bfqxz.

I hope this is of some use.

Signed winger.

3. Rourke Decker on February 5, 2012 at 8:47 am said:

Tripled letters remain rare in English, but they are surprisingly common in German, especially since the recent Rechtschreibungsreform. The words Schifffahrt (“boat trip”) and Stillleben (“still life”) come immediately to mind, although there are many others.

4. Rourke, your comment provoked my curiosity, so I did a quick search:

http://oranchak.com/german-triples.txt

5. Rourke: tripled letters in English? Nothing could be freeer! 🙂

6. Rourke Decker on February 5, 2012 at 7:15 pm said:

That is an amazing list, David, and quite eye opening. Thank you for taking the time to compile it, even if it were through automated means. 😉

Unfortunately, it’s in plain-text ASCII, so the Eszett (ß) and umlauted letters (ä, ö, ü) don’t show up correctly. It would be nice if the list could be reformatted in Unicode so that the words would appear in all their glory. A picayune gripe, I know, but what can I say? I’m a nerd.

Nick, it would not surprise me if tripled consonants were more common in compounding languages, of which the Germanic languages are prime examples. Extremely long words comprised of compounded shorter words are possible, though less frequently used today in the past. In the 2009 German novel I am currently in the process of translating appears the following beautiful example of what is possible in such languages: a pressure washer is described as a Kulturpflanzenbegleitervernichtungsapparat (“crop plant escort extermination device”). “Crop plant escort” is a nice German euphemism for “weed.” A famous Swedish example is produktionsstyrningssystemsprogramvaruuppdatering (“production controller system software update”).

7. i think it is a Poetry about animals

8. Anne-Lise Pasch on January 28, 2013 at 9:02 am said:

If you remove all the nulls (assuming all the positions with a frequency of 3 or less are null – as these ONLY occur in column 14), you’re left with a 13 character cipher alphabet (exactly half of a normal alphabet). Is there a cipher that uses only 13 characters (possibly representing digrams?)

On another tack, if this is a columnar transposition, on a rotated 14×14 grid, and the nulls occur only in the last column (last row when you rotate for deciphering), when you transpose, all the nulls should logically remain grouped, as this would be the padding at the end of the plaintext on the last row (using null characters, like z, etc.) So, if we group those columns together and ignore them (initially) for an attack, we’re only dealing with a 7-wide column attack space to find anagrammed trigrams within. And if we use the polybius frequencies for arranging common letters in the cipher alphabet, we should also have a much reduced set of initial cipher alphabets, and only use the top 20, say, trigrams to search for (to arrange columns with for scoring), this should allow for a brute force attack…

Or am I going crazy?
Of course, all this hinges on the number of potential nulls, and again I’m making assumptions. This cipher’s got me thinking in circles. :>

9. jake on April 5, 2013 at 7:51 pm said:

comment on the triple letters: What I have read “decoders” are looking for words with 3 letters in a row. But, what if it is two words: ie ‘ah but stiLL Love will last’ note the 3 Ls in a row. Something else I have noticed that ever other set of 5 numbers, is a large number then smaller number then larger number…. also that there are no double numbers as 11,22,33,44… also I believe that the 92000 at the end is not just null value but rather a special character (period or question mark).
Just rambling thoughts of an old man… jake

10. william heathway on February 23, 2015 at 1:35 pm said:

I think 75628 and 28591 says TNT away that’s all I will give away if I am right

11. You solve with a 10 by 10 grid. Common letters like t are placed several times to avoid letter frequency.

12. SirHubert on February 24, 2015 at 12:10 pm said:

xplor: and the solution is…?

13. Jack Forbes on July 13, 2015 at 7:22 am said:

Anne-Lise Pasch, how do you know so much about cryptology? What is your background in this? A hobby or professional?