As far as I can tell (from reading the Wikipedia article about it) decrypting a message encrypted with the vigenère cipher is possible on paper.

But what would happen if the message is encrypted multiple times with a set of different keywords?

"Thwaites challenged Babbage to break his cipher encoded twice, with keys of different length. Babbage succeeded in decrypting a sample, which turned out to be the poem "The Vision of Sin", by Alfred Tennyson, encrypted according to the keyword "Emily", the first name of Tennyson's wife. Babbage never explained the method he used. Studies of Babbage's notes reveal that he had used the method later published by Kasiski, and suggest that he had been using the method as early as 1846." Link to the Kasiski examination

(I must admit, I don't quite understand how to apply the method mentioned above effectively).

An example message:

First time (keyword: "ASIMPLEPASSWORD")
This is a secret message that was encrypted multiple times with a vigenere cipher

Second time (keyword: "RANDOM")

Third time (keyword: "UNKNOWNTOYOU")


The reason I'm asking is simple research (and curiosity). I'm writing a story involving encrypted messages in a fantasy setting (so, no computers!). In-depth knowledge about the cipher (including methods and formulas known today) can be used, though.

I'ld like to see if this method would be impossible to crack by "classic means/hand", or if it has it's own (perhaps obvious to experts) weakness.

How would an expert in this field approach the problem?

At what point would that expert either realize the message was encoded multiple times or mistakingly believe it's not a vigenère cipher?

What additional hint or information would that expert need to decrypt the message?

  • $\begingroup$ I found that if you decrypt a vigenere message and you know all the keys used, the order in which you decrypt them doesn't matter. I used the website Dcode(dcode.fr/vigenere-cipher) to prove this thought but I wasn't sure if it was a law. I've used 2 keys and 3 keys but the order was switched during the decoding process and the message was the same. $\endgroup$
    – Neo1009
    Commented Oct 2, 2018 at 15:53

1 Answer 1


My impression is that it would at most take a longer ciphertext to be decrypted. Imagine first the scenario where the different keys have the same length:



In that scenario, the keys add up and the result of applying $E_{K_2}(E_{K_1}(m))$ can be expressed as one single operation: $E_{K*}(m)$, where $K*=dscwr$. Therefore, not a lot is two encryptions are nested with keys of the same length.

If you use keys of different lengths, you'll meet the same combination of letters at some point. Let's say you have a key of length 3 and a key of length five:

$K_3=abc$ $K_5=rolex$

After the position 15 ($5 \cdot 3$), you'll encrypt again a value with the same combination as in the position 1. Therefore, it is as if you had two 15-character keys:



So, for this case in particular, you win nothing.

Now, for the most general case, keep in mind that no paper cipher is a serious rival against the existent computational power. However, if by some clever narrative you manage to limit the powers of the adversary to just paper based attacks, there are paper algorithms much stronger than Vigenère. You can start by taking a look at Solitaire: https://www.schneier.com/cryptography/solitaire/

  • $\begingroup$ I must admit I absolutely overestimated (multiple uses of) Vigenère, since I thought using it multiple times would make things more difficult. But I can work with that, now that I know the problem - I can't make the cipher too strong since it's actually required for the story that it's eventually cracked. Thanks for your help! $\endgroup$
    – Katai
    Commented Mar 20, 2016 at 19:25
  • $\begingroup$ @Katai It is harder if the keys are words instead of randomized sequences. With "hello" and "world" you can easily guess characters of the key and check if the key is correct. For "dscwr" this is not the case (unless the adversary thinks it is shorthand for "discwear" of course :P ). For classical ciphers a fully random key is less likely to be used. $\endgroup$
    – Maarten Bodewes
    Commented Oct 2, 2018 at 16:51

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