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| visits | member for | 1 year, 9 months |
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| stats | profile views | 105 |
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Apr 27 |
answered | Cipher Feedback Mode |
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Apr 26 |
comment |
Can RSA encryption produce collisions? @Ashwin: Well, an outline of a proof would look like: if $GCD(e, p-1)=1$, and if $m_1 \neq m_2 \mod p$, then $m_1^e \neq m_2^e \mod p$ (note: the proof of this relies on the primality of $p$). And, by symmetry, if $GCD(e, q-1)=1$, and if $m_1 \neq m_2 \mod q$, then $m_1^e \neq m_2^e \mod q$. Now, if we combine these two statements using the Chinese Remainder Theorem, we get: if $GCD(e, lcm(p-1, q-1))=1$ and if $m_1 \neq \m_2 \mod pq$, then $m_1^e \neq m_2^e \mod pq$. Take the converse of that statement, and that's the statement you're asking about |
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Apr 22 |
answered | How secure is the Vigenère cipher in file encryption if you encrypt the password first? |
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Apr 22 |
revised |
How secure is the Vigenère cipher in file encryption if you encrypt the password first? edited tags |
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Apr 20 |
reviewed | Approve suggested edit on Proving item association without revealing one of the associated items |
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Apr 20 |
answered | Distributed knowledge problem |
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Apr 20 |
comment |
Distributed knowledge problem There's something missing in this problem statement; you said that each card represents one of the movies; what does that actually mean? Does it mean that Alice and Bob can jointly open one of the cards? If so, how much trust at this point do you put in Alice and Bob to be honest during the reveal? Also, do you care if, say, Bob knows that a particular card is from Alice, or does that need to be hidden as well? |
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Apr 20 |
comment |
Proving item association without revealing one of the associated items Yes, one the the basic assumptions (Collision Resistance) of secure hash functions is that it is infeasible to find two different values X, Y with Hash(X) = Hash(Y) |
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Apr 20 |
comment |
Proving item association without revealing one of the associated items Well, as long as the R value is selected randomly, and has enough possible values that searching through them is infeasible, that should be fine. |
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Apr 20 |
answered | Proving item association without revealing one of the associated items |
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Apr 20 |
revised |
How to solve MixColumns Gave explicit evaluation of the example the submitter gave |
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Apr 19 |
revised |
How to solve MixColumns Extended the answer to cover the next thing the OP is likely to stumble against |
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Apr 19 |
answered | How to solve MixColumns |
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Apr 19 |
reviewed | Approve suggested edit on hmac tag wiki excerpt |
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Apr 19 |
reviewed | Approve suggested edit on hmac tag wiki |
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Apr 18 |
revised |
How much bigger does a precomputed lookup table get when salt is added? edited body |
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Apr 18 |
comment |
Can RSA encryption produce collisions? @HenrickHellström: from the 'nits-r-us' department, the example $e=3$, $N=91$, $p=7$, $q=13$, $m_1=5$, $m_2=6$ shows $e|p-1$ and $m_1^e \equiv m_2^e (\bmod N)$ but $m_1 \neq m_2 (\bmod q)$. On the other hand, if you add the condition $gcd(e, q-1) = 1$, then your statement is true. |
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Apr 17 |
answered | How much bigger does a precomputed lookup table get when salt is added? |
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Apr 17 |
comment |
Security equivalence proofs for breaking RSA Actually, no, that paper doesn't claim that 'if you can break RSA (that is, find the $e$'th roots modulo a composite of unknown factorization), you can factor the composite.' Instead, the paper shows that, given $N$, $e$ and $d$, you can factor $N$ (which we knew already with the probabilistic algorithm). The paper does not show that 'if you could break RSA, you can recover $d$'. |
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Apr 17 |
comment |
Most frequently used digital signature schemes in the recent years In my experience, the signature scheme of choice if you want to minimize the signature size is ECDSA. I personally haven't seen DSA used since the RSA patent expired. |
