Ilmari Karonen
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 15h comment AES128 Decryption in Objective-C just slightly off @Journeyman: In CBC mode, the IV is simply XORed with the first block of plaintext before encryption / after decryption. (This is also why it's so important for security that the IV be random and unpredictable in CBC encryption.) So if you know the plaintext you should be getting, you can just XOR that with the plaintext you are getting, and then XOR that with the IV you're using to obtain the correct IV. 2d comment Steganographic embedding of an RSA-encrypted message into an image This question has nothing to do with RSA specifically. Your real problem seems to be that you haven't understood how LSB steganography works. Feb 7 comment Common Modulus Attack not reproducible Um, no you can't. By definition, if $\gcd(m,n) = k$, then any multiple of $m$ modulo $n$ is also a multiple of $k$. The closest you can get is finding a pseudoinverse $m^*$ such that $m \times m^* = k \pmod n$ (and you can do that with the same extended Euclidean algorithm used to find normal modular inverses). Feb 4 comment Find the minimum Adversary's Advantage @stathisb: Are you missing the definition of advantage for MAC schemes, or the way to obtain a non-negligible advantage for this particular question (or something else)? If the former, your textbook (or lecture notes) should have it, assuming that it's using the term "advantage" for MACs in the first place; if the latter, see my answer below. Feb 4 comment Find the minimum Adversary's Advantage @stathisb: ...where you've now introduced three new undefined symbols ($K$, $M$ and $T$). I can guess what those mean (I bet they're the keyspace, message space and tag space respectively), but such notation isn't really unambiguous or universal enough that you could just assume that everybody will understand it (they way in math you can usually assume e.g. that $\pi \approx 3.14159265$), especially if they learned crypto from a different textbook than yours. But anyway, more important than all this notation is really that you still haven't told us what you've tried and where you're stuck. Feb 4 comment Find the minimum Adversary's Advantage @RickyDemer: I suspect it just means that they're representing their MAC scheme as a pair of functions $S$ and $V$, which probably stand for "signer" and "verifier". Wikipedia uses a similar notation. But yeah, the question should really mention that, or at least link to a definition. Feb 4 comment Security of a parallelizable block cipher mode Also, this construction looks vaguely similar to OCB mode. Are you familiar with that? Feb 4 comment Security of a parallelizable block cipher mode Maybe a silly question, but what's $\boxplus$? I assume $\oplus$ means bitwise XOR, but I don't really know a standard meaning for $\boxplus$. Feb 1 comment SHA1 collisions and the impact for ECDSA signatures If the attacker only has control over $m'$, then they need a (second) preimage attack to forge a signature. But there are practical scenarios where both $m$ and $m'$ may be controlled by the attacker, in which case a collision attack suffices. For example, the MD5 collision attack was used to forge SSL certificates by getting an established CA to sign an innocent-looking certificate, and then replacing it with another certificate with the same MD5 hash, but far higher privileges. Feb 1 comment Can an AES-256 key be generated for a large number of cribs? We seem to be getting questions like this one with some regularity. Alas, the answer is always the same: if the virus writer didn't screw up somehow (and didn't get arrested and forced to surrender their private keys), you can't break the encryption. Jan 25 comment Proving that a node is the only child Are we allowed to update the hashes of existing nodes when new nodes are added? (I assume yes, since the problem is obviously unsolvable otherwise, but it would be good to state this explicitly in your question.) And are there any constraints on which hashes we may update (e.g. can we update the hashes of the added node's siblings, or only the hash of its parent, or...)? Jan 23 comment Complexity using RSA Not off the top of my head, because I don't have the "standard exponentiation algorithm" memorized. Also, now that I think about it, are we allowed to choose $e$ freely? Using $e=3$ should make the exercise simpler and also allow a higher $m$ than using a random \$e