In BBC’s Sherlock, Season 1, Episode 2 “The Blind Banker”, the Black Lotus crime syndicate uses ancient Chinese number characters to refer to pages and words in a commonly known book (which was currently the A-Z London Guide).

What class of cryptographic functions would this most closely fall under? Is it symmetric/private key encryption? Or asymmetric/public key encryption? Or something else?

My understanding is that the key is the common book, and the Chinese characters are the ciphertext of the original words. Hence this would make it symmetric encryption? But at the same time, the book is publicly accessible so it's not strictly a secret/private key.


1 Answer 1


This sounds like a straight forward book cipher.

Symmetric encryption implies that the knowledge of the key is sufficient for both encryption/decryption.

Asymmetric encryption implies that the ability to perform the encryption/decryption operation is separated into two distinct keys.

But at the same time, the book is publically accessible so it's not strictly a secret/private key.

Which book is used may be the secret key. Also, which book is used could be public knowledge, but certain subsections of it are the key.

Another way to look at it is that the set of possible bit strings of a given size is public knowledge. But if you select one particular bit string secretly from this publicly known set, it does not mean your secret key is then public information.

  • 2
    $\begingroup$ The secret key is which book the cipher uses. $\endgroup$
    – Meir Maor
    Commented Jan 28, 2018 at 18:18
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    $\begingroup$ Buy yourself a copy of David Kahn’s “The Codebreakers” it’s all covered in there $\endgroup$
    – Stevetech
    Commented Jan 28, 2018 at 19:51
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    $\begingroup$ Also, Benedict is totally wrong about "modern code breaking method won't work", as a HPC cloud digital library would be capable of scanning through the entire British Library to find a few books that could potentially have been the key, which given the limitedness of human history, I suspect it's doable in a complexity not too greater than breaking DES. $\endgroup$
    – DannyNiu
    Commented Jan 29, 2018 at 11:42

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