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| visits | member for | 1 year, 9 months |
| seen | 2 hours ago | |
| stats | profile views | 105 |
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11h |
comment |
How random are comercial TRNGS @user1028028: actually, it's not true that a determanistic PRNG must be periodic; it's easy to make a nonperiodic PRNG by expanding the state over time (and besides, we don't care about repeats after (say) 2^256 outputs; we never come even close to that; and we usually reseed the PRNG periodically anyways). In any case, fgrieu answered why postprocessing the TRNG output through a PRNG is useful, in his experience (and mine, however my experience is more modest), "the unconditioned output of a TRNG source is never free from some detectable bias". Is that contrary to your experience? |
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1d |
comment |
What is the fastest elliptic curve operation f(P) in affine coordinates such that f^n(P)=P only if n is large? The standard way to generate random points is to select a random value for X, check to see if there's a solution for the elliptic curve equation with that value, and if these is, pick one of the two possible values for Y. Or, do you need random values with known relationships, or for which you can compute output number N+1 given output number N? |
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1d |
reviewed | Approve suggested edit on Decrypting the Encrypted hex applcation data with encryption keys |
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1d |
reviewed | Approve suggested edit on What is the fastest elliptic curve operation f(P) in affine coordinates such that f^n(P)=P only if n is large? |
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2d |
comment |
Is the likelihood of a birthday collision linear (evenly distributed) for all ciphers? When are they not? Well, if you apply the birthday principal to an nonuniform distribution, collisions become more likely than over a uniform distribution over the same number of elements (assuming, of course, that all tries use the same distribution). So, it might make some attack more difficult on average, however it's not at all clear what those attacks are actually under discussion. If the attack relies finding collisions, a nonuniform distribution would appear to make things worse, that is, easier to perform, and hence less secure. |
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May 20 |
awarded | Enlightened |
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May 20 |
awarded | Nice Answer |
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May 20 |
revised |
Physical Level Encryption Expanded explination |
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May 20 |
answered | Physical Level Encryption |
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May 16 |
answered | How does OAEP improve the security of RSA? |
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May 16 |
comment |
Generating IV in TLS 1.2 @D.W.: why, yes, I do have a proof of security: see eprint.iacr.org/2008/121.pdf ; see the part where they show an encrypted counter for a CBC mode IV is secure |
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May 15 |
comment |
Generating IV in TLS 1.2 @user1966074: The decryptor side does the standard CBC-mode decryption, which is (for TLS) precisely as you stated. |
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May 15 |
reviewed | Approve suggested edit on Partial collisions for md5 |
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May 15 |
answered | Generating IV in TLS 1.2 |
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May 14 |
comment |
Encrypting a broadcast channel What's the security goal behind having multiple keys? So that we can encrypt data that can be decrypted by a specific subset of the subscribers (e.g. Alice can decrypt it, but Bob can't, even though both Alice and Bob are both subscribers)? Alternatively, is it so that you can do traitor-tracing (that is, if someone leaks his key, we can figure out who that is)? |
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May 14 |
reviewed | Approve suggested edit on Why is H(k||x) not a secure MAC construction? |
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May 9 |
comment |
Increased CRC collision probability when adding bits to input message @jmbeck: hmmmm, would it be possible to encode the version bits in the message length? That is, version 1 messages are a specific length (which is what you're currently doing now), version 2 messages are longer (and have an explicit version field, a longer check value, etc). I don't know if that's possible; I thought it may be worth considering. |
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May 9 |
awarded | Enlightened |
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May 9 |
awarded | Nice Answer |
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May 8 |
answered | Increased CRC collision probability when adding bits to input message |
