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    SubjectRe: Using compression before encryption in device-mapper
    Date
    Feel free to ignore all of my reply; please note that I am not trying to "be 
    right" or to "be wrong" but I still do not understand ... Thank you still for
    your time and pedagogy.

    > Yeah, sure, the attacker has no idea what the plaintext of those
    > blocks is, but if they appear often enough, it has to be something
    > quite common. Something like, say, all ones or all zeros. Or like
    > one of those 48 common huffman encodings thereof.
    > [...]
    > So what! You end up with maybe three bits per zero (assuming all
    > zeros). Depending on the size of random data up front, they start
    > with bit 1, 2 or 3. Makes 3*2^3 or 24 possibilities. Same for all
    > ones, give a total of 48. Great, a dictionary attack is 48x slower
    > now!
    > [...]
    > Still, towards the end of all-ones or all-zeros, each byte will be
    > encoded with the same 1-3bit value.
    The point I fail to understand is the following : you know the enciphered
    value of these 1-3bits. But how can you know what is
    compressed-but-deciphered 1-3bit value ? Ok my text contains only 0s. OK
    these 0s appear to be "011" once encrypted. How do you launch your
    dictionnary attack ? You do _not_ (?) know what the 3bit deciphered code for
    "0" is. Or maybe you do ?

    > [...]
    > In that case, what's your point. If the key is strong and the
    > encryption is strong (I sure hope, AES is), nothing short of brute
    > force can be successful. What are you protecting against?
    Maybe my "endless" story is absurd, but I am _not_ protecting against weak
    keys; I am trying to protected against weak _data_ , which is the basis for
    dictionnary attacks even in the case of perfectly random keys.

    Thank you for having read till here,
    Guillaume.

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