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I read about XEX mode of operation in Wikipedia My question is:

There are two keys that are XORed against plaintext, if instead of two XORed keys I add 4 or 6, does it increase the security of block cipher mode?

In the case of 4 XOR operations, I mean XOR the first key, XOR the second, encrypt, and XOR the third and fourth key.

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The xor–encrypt–xor (XEX) simply is;

$$C = E_k(k_1 \oplus P) \oplus k_2$$

now if you x-or with 3 other random keys one will get

$$C = E_{k \oplus k_3}\left((k_1 \oplus k_4) \oplus P\right) \oplus (k_2 \oplus k_5)$$

  • In the attacker's sense, this doesn't matter, and

  • For the improvement of the security of the scheme, this is not an improvement since the security level is the same;

    Once we have 3 keys $k,k_1,k_2$ and now we again three keys, too;

    • $k' = k \oplus k_3$
    • $k_1' = k_1 \oplus k_4$
    • $k_2' = k_2 \oplus k_5$

    Therefore we have new 3 keys with the same key sizes.

If you want to increase the security just use 256-bit encryption, which will be enough to cover pre-and post-quantum adversaries. Increasing the size of the encryption key $k$ doesn't mean that you can increase $k_1$ and $k_2$ since they are bounded by the block size that is generally 128-bit.

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    $\begingroup$ I undestand, thank you for the answer. $\endgroup$ Jan 14 at 0:38

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