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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
answered Find the minimum Adversary's Advantage
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
reviewed Close Compression in Symmetric-Encryption?
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
28
revised How can a message encrypted with the public key be decrypted with the private key?
clarify title; misc. copyedits; remove "thanks" per SE style guide
Jan
27
revised How can a message encrypted with the public key be decrypted with the private key?
added 2 characters in body; added 1 character in body
Jan
27
revised How can a message encrypted with the public key be decrypted with the private key?
added 2 characters in body
Jan
27
answered How can a message encrypted with the public key be decrypted with the private key?
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 $e<n$. (Also, the constant factor 100 in the assumed cost of multiplication seems pretty high, at least for multiplication by a small constant.)
Jan
20
comment Complexity using RSA
For bonus points, mention that you would never use RSA like this in practice, because it's both inefficient and insecure.
Jan
18
comment Compromise between HMAC and Digital Signature, by encrypting and sending secret key?
@user29944: To be precise, I meant to say that the time needed to hash the message is strictly less than the time to HMAC (or sign) it, simply because hashing the message is one step in both HMAC and in signing. Typically, the full signing process will be somewhat slower than HMAC; but the difference is by a constant amount (depending only on the signature and hash algorithms, and on the signature key length), and thus becomes relatively insignificant for sufficiently long messages.
Jan
12
awarded  Enlightened
Jan
12
awarded  Nice Answer
Jan
12
revised Could this “symmetric RSA” scheme provide key compromise resistant communications?
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