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If I encrypt a 1MiB file with AES-CBC (or any other cipher) and XOR a 128KiB of (truly) unpredictable random data repeating until the end of file, will I have a security of 1048576-bits (128KiB*8)?

This question is a little weird, but I would like to know if this scheme has a security flaw (maybe known-plaintext attacks).

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This will be only as secure as AES-CBC, as the repeating XOR you're describing is massively vulnerable to a myriad of attacks, including known-plaintext attacks. Also remember that you could "cancel out" the 128 KiB by XORing two blocks together, since $P_1 \oplus K \oplus P_2 \oplus K = P_1 \oplus P_2$.

Don't try to chase large key sizes. 256 bits of key material is more than enough.

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  • $\begingroup$ Yes, I was trying to chase large key sizes, pardon me. =) $\endgroup$ Nov 28, 2022 at 0:57
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    $\begingroup$ @phantomcraft There's really no need to do that. 256 bits is plenty, although you can use XTS to get a little more strength for "free" (384 bits when you take into account meet-in-the-middle). $\endgroup$
    – forest
    Nov 28, 2022 at 0:58
  • $\begingroup$ In a quantum scenario I would get 192-bits with AES-256-XTS, am I right? $\endgroup$ Nov 28, 2022 at 1:06
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    $\begingroup$ @phantomcraft Well, sort of, but see crypto.stackexchange.com/a/102672/54184. While it would reduce it to 192 bits, quantum computers simply don't scale when running Grover's algorithm. Even 256 bits (thus 128 vs Grover's) is more than enough. $\endgroup$
    – forest
    Nov 28, 2022 at 1:08
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    $\begingroup$ They can, but $2^{128}$ quantum operations is way more difficult to achieve than $2^{128}$ classical operations. $\endgroup$
    – forest
    Nov 28, 2022 at 1:15

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