2
$\begingroup$

I’m reading “The Security of Multiple Encryption in the Ideal Cipher Model” and I’m having a hard time understanding something. How do I calculate the bounds of cascade cipher from the following formula (where $k$ determines keysize and $n$ is the blocksize)?

$\exp\left(k+\min\left\{\frac{k(ℓ′−2)}{2}, \frac{n(ℓ′−2)}{ℓ′}\right\}\right)$   where $ℓ$ is the cascade length and $ℓ > 1$

If there is a cascade of AES, Blowfish and DES then what will be the value of $k$? and what will be the value of $n$? is this just addition of the key size used by these block ciphers for determining the value $k$ and addition of the blocksizes for determining the value of $n$?

If the case is AES $\to$ DES $\to$ AES $\to$ DES $\to$ Blowfish, then is this just five times addition to determine $k$ and $n$?

$\endgroup$
2
$\begingroup$

It seems the result is specifically for multiple encryption with a single cipher (like in 3DES). It probably applies for different ciphers as well, but key and block sizes would need to be equal. You might get a lower bound by using the minimum key size, but don't quote me on that.

However, I don't think this is really relevant for a practical implementation. AES alone has a large enough key space that brute force attacks on the key are impossible. (Assume AES-256 if you care about quantum computers or are very optimistic about Moore's law.)

The formula tells you nothing about what the security is if some of the ciphers in your cascade are broken, which is the only reason you would use different ciphers in the first place. Therefore, it tells you nothing very important about the security of the cascade in practice, unless all the ciphers you used would be insecure on their own (again, 3DES). The brute force resistance is simply "enough".

$\endgroup$
  • $\begingroup$ I understand what you are trying to say. $\endgroup$ – Giliweed Sep 21 '14 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.