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Re: Quor's cipher
[email protected] (Matt Ghio) writes:
> What about this:
> If (a+b)^(a0+b0) == 0, then the plaintext is the same as the ciphertext.
> This happens for one out of every 256 bytes. Ordinarilly this isn't a
> problem, but if the key is reused, and there is no IV, it can leak a byte
> of plaintext.
> So it seems that you would need to change the key for each message, or at
> least use a random initialization vector.
How are you planning to detect which bytes are passed in this way ?
Chosen plaintext attacks would do it, and show where (a+b)^(a0+b0) == 0.
Looks like you've just doubled our progress.
If the key is reused with a different message I don't think there's a
weakness. An IV is a good idea, but aren't we _attacking_ this thing ?
I've grabbed a few search-engine hits and not read them yet. I'll be
looking for clues there.
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