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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 7 months |
| seen | Jan 10 at 13:25 | |
| stats | profile views | 6 |
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Nov 8 |
comment |
How can I use eulers totient and the chinese remainder theorem for modular exponentiation? To do the computations, you use what is the most efficient: -1 or 4. For instance, I'd compute the $5^7 \mod 11$ above as $5^{-3} \mod 11$ instead. (As $5^3=4\mod 11$ and $4\cdot 3=1\mod 11$, $4^{-1}=3\mod 11$.) For a mathematician, there's an equivalence class, the label does not really matter, although common practice is indeed to use non negative numbers. |
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Nov 8 |
comment |
Randomized algorithms and the one time pad The standard naming, coined by Goldwasser and Micali, is probabilistic encryption. |
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Nov 8 |
awarded | Enthusiast |
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Nov 3 |
comment |
Why does HOTP use such a complex truncate function? You meant from $2^{80}$ to $2^{16}$. |
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Nov 2 |
answered | “Padless” One-time-Pad encryption |
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Oct 30 |
revised |
How can I prove in zero knowldege that an ElGamal shuffle is correct for a special setting? fixed the title |
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Oct 30 |
suggested | suggested edit on How can I prove in zero knowldege that an ElGamal shuffle is correct for a special setting? |
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Oct 30 |
comment |
Modifications of CBC-MAC A message of length bigger than $2^n$ does not exist :D This is forbidden by the format constraint. (You've to understand that the actual limit value is not crucial: you could also take a limit of $2^{2n}$ and code the bit length over two blocks, this would achieve the same effect.) |
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Oct 29 |
comment |
Modifications of CBC-MAC @Avery: the constraint on the message format prevents a valid message from being extended into another valid message by a simple append. |
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Oct 29 |
revised |
Modifications of CBC-MAC fixed a typo and made the size computation more explicit |
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Oct 29 |
comment |
Can a proof be constructed to show there is no distinguisher? The hash function Skein is not exactly what I would call "simple" from the point of view of understanding the way it maps its inputs to outputs. Otherwise, one would "simply" find pre-images and collisions for it! |
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Oct 29 |
revised |
Does chaining random number generators lead to loss of randomness? fixed typo in the title |
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Oct 29 |
suggested | suggested edit on Does chaining random number generators lead to loss of randomness? |
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Oct 29 |
comment |
Modifications of CBC-MAC It is correct that any $\rho\neq0$ will do. However, there was a reason: the adversary in 1) cannot be distinguished from the real signer and that it is straightforward to show it. (Of course, it might also be the case with other ways to choose $\rho$, but harder to prove.) Note that it was never stated in the question that the adversary has to be indistiguishable from the legitimate signer, but this is a nice property (from an attacker point of view) as it shows that forgeries cannot be detected. |
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Oct 28 |
answered | Modifications of CBC-MAC |
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Oct 28 |
answered | What is the complexity of the Square attack against the reduced 4-rounds 128-bit Rijndael variant? |
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Oct 28 |
comment |
Why do we need in RSA the modulus to be product of 2 primes? @fgrieu: nice link. It might be worth noting that the encryption scheme also "works" with $N=p$ a big prime, but then becomes symmetric (and was actually invented by Polhig and Hellman) as opposed to RSA which is an asymmetric one. |
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Oct 27 |
revised |
Subgroups generators with respect to group generators of composite order some simple formating and remove prime for the composite number $N$ |
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Oct 27 |
answered | Does chaining random number generators lead to loss of randomness? |
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Oct 27 |
suggested | suggested edit on Subgroups generators with respect to group generators of composite order |
