Security Corner

Feb 29 2008   2:32AM GMT

The Unsolved D’Agapeyeff Cipher

Ken Harthun Ken Harthun Profile: Ken Harthun

Sometimes, it’s a good thing to take a breather from the routine, to venture off into something more fun than the serious day-to-day concerns of network and computer security. One of my interests is cryptography, especially its history, and I love to play around with cryptograms in the daily newspaper, even though they’re just simple substitution ciphers (though there are some puzzle books out there that use polyalphabetic and transposition ciphers).

There’s no question that computers have taken cryptography well out of the realm of human-generated codes and ciphers. Done properly, modern encryption systems produce output that appears to be nothing more than random noise to a human–and no human will ever be able to break those ciphertexts without the help of powerful computers. Yet, there are human-generated ciphers that haven’t been cracked. One of those is the D’Agapeyeff cipher, which appears as “…a cryptogram upon which the reader is invited to test his skill” in the first edition of “Codes & Ciphers, ” written by Alexander D’Agapeyeff, published by Oxford University Press in April, 1939.

The book is an elementary text on classic encryption methods and the cryptogram is placed on the final page of the final chapter which details methods of decryption of the various types of ciphers. Here’s the cryptogram as it appears in the book (this was omitted from later editions for reasons unkown):

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

I assumed (correctly, I think–see this article) that two numbers represent one letter and that this was some sort of simple substitution cipher. I divided the cryptogram thus, omitting the three zeros that are obviously nulls:

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

You can see that no pair begins with a number less than six and no pair ends with a number greater than five. This suggests a matrix like this:

1 2 3 4 5
6a b c d e
7
8
9
0

Using this hypothetical grid, 61 is “a,” 65 is “e,” etc. That’s as far as I’ve managed to go.

Anyone else like to play with this?

Cheers!
The Geek

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  • Popcorn
    Fourth line, third from the end: 04.
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