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Aug
27
comment What's the point of a Meet In The Middle attack using i.e. double AES with throw-away keys?
@maincodebase: not quite any such attack; if the attacker has enough partial information to limit the number of potential messages to a small bound, as in "The first letter of my password is X", then an AON transform won't help you. On the other hand, we have ciphers that are immune to known plaintext (and chosen plaintext) attacks; hence we don't worry too much about how we might work with a cipher that isn't.
Aug
27
answered What's the point of a Meet In The Middle attack using i.e. double AES with throw-away keys?
Aug
26
answered Are variable-length crypto hash functions still susceptible to collisions?
Aug
26
reviewed Leave Open Good challenges for a crypto competition for teenagers
Aug
26
reviewed Close What's the best packet cipher mode for use with UDP for example? Advantages or disadvantages of any alternatives?
Aug
26
reviewed Close Counter Mode: how to choose the nonce part of the counter?
Aug
25
comment Found a way to crack AES-128, what now?
@hunter: it would count as a break even if he could not decrypt a single ciphertext; it would count as an extremely bad break even if he needed a few gigabytes of chosen plaintext.
Aug
25
comment Solution with high decryption cost and low encryption cost
@otus: I don't know why we'd want to slow down the decryptor either; however I believe that's what devd asked for...
Aug
25
comment Solution with high decryption cost and low encryption cost
@otus: If you are saying "he can decrypt multiple messages quickly", well, no, that's why I stir in $S$ (a nonce) into the puzzle that hides $r_i$; different messages will have different $S$ values. If you're saying "he can then test multiple values of $Key$ quickly", I didn't believe that was an issue; $Key$ is (say) a 128 bit value (and hence is immune from brute force search). $Key$ is present solely in case we want to forbid a third party from decrypting at all. If that is not a requirement, $Key$ can be omitted.
Aug
25
answered Solution with high decryption cost and low encryption cost
Aug
25
reviewed No Action Needed Found a way to crack AES-128, what now?
Aug
25
reviewed Close Seemingly simple decryption question
Aug
25
reviewed Leave Open Solution with high decryption cost and low encryption cost
Aug
25
reviewed Close Variants of AES?
Aug
25
reviewed Close Efficiently map $2^n$ unique 64-bit vectors to $2^n$ $n$-bit vectors where $n < 64$?
Aug
25
comment Efficiently map $2^n$ unique 64-bit vectors to $2^n$ $n$-bit vectors where $n < 64$?
I'll vote to close as well; but you might want to consider: "what properties do you want from your mapping?" If you want just any mapping, just chopping off $64-n$ bits off the vectors works fine -- is there a reason why that isn't sufficient?
Aug
24
comment Simple proof that shows AES is not a uniform permutation on any n-bit string?
@ddddavidee: actually, there are block ciphers that can implement an odd permutation; they're not common, but they most certainly do exist. AES (as well as the vast majority of block ciphers) will always do an even permutation.
Aug
23
awarded  modes-of-operation
Aug
21
comment CBC-MAC key can be added to ciphertext?
@Kingland: yes (or somehow otherwise have both sides arrive at the same secret keys)
Aug
21
revised CBC-MAC key can be added to ciphertext?
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