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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year, 10 months |
| seen | 16 mins ago | |
| stats | profile views | 106 |
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Mar 9 |
comment |
Shannon entropy calculation: is $H(A|R·A) = H(A)$? @user996522: It is a group because if we consider the set of all invertible matrices and consider the operation of matrix multiplication, it fulfills all the requirements of a group; it is closed (if A and B are invertible matrices, so is $A\cdot B$), it has associativity, it has an identity element, every element has an inverse). Why being a group is important is because, for any group operation $\oplus$, if R is a uniformly and independently distributed group element, then so is $R\oplus A$, in particular, $R\oplus A$ is independent of A. |
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Mar 8 |
revised |
Shannon entropy calculation: is $H(A|R·A) = H(A)$? Nit: I noticed that both matrices were singular if $n | 3$ |
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Mar 8 |
revised |
Shannon entropy calculation: is $H(A|R·A) = H(A)$? Amplified the last paragram a bit |
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Mar 8 |
revised |
Shannon entropy calculation: is $H(A|R·A) = H(A)$? Totally misunderstood the question at first |
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Mar 8 |
answered | Shannon entropy calculation: is $H(A|R·A) = H(A)$? |
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Mar 7 |
reviewed | Approve suggested edit on preimage-resistance tag wiki excerpt |
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Mar 7 |
reviewed | Approve suggested edit on preimage-resistance tag wiki |
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Mar 7 |
reviewed | Approve suggested edit on collision-resistance tag wiki excerpt |
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Mar 7 |
reviewed | Approve suggested edit on 2nd-preimage-resistance tag wiki excerpt |
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Mar 7 |
reviewed | Approve suggested edit on collision-resistance tag wiki |
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Mar 7 |
reviewed | Approve suggested edit on 2nd-preimage-resistance tag wiki |
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Mar 6 |
comment |
reverse of md5sum @Marcos: it sounds like you're looking for a data compression method. Well, you're pretty constrained in what you can expect. If you expect the exact same preimage, well, if you have $2^N$ possible preimages, then the lengths of the hash and the seed must total at least $N$ bits long. If you expect a similar preimage, well, how well you can do is highly dependent on your definition of "similar". |
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Mar 6 |
answered | reverse of md5sum |
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Mar 6 |
comment |
Is it safer to generate your own Diffie-Hellman primes or to use those defined in RFC 3526? Actually, it's not actually true that "it doesn't matter what prime you use"; certain primes (say, primes where $p-1$ is smooth) are a really bad idea. In addition, it's a good to generate $p$ so that you know a large prime factor $q$, so that you can generate a generator for a subgroup that size. |
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Mar 6 |
answered | Is it safer to generate your own Diffie-Hellman primes or to use those defined in RFC 3526? |
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Mar 6 |
revised |
Is it safer to generate your own Diffie-Hellman primes or to use those defined in RFC 3526? edited title |
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Mar 3 |
answered | What should be the size of a Diffie-Hellman private key? |
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Mar 2 |
answered | What should be the size of a Diffie-Hellman private key? |
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Mar 2 |
comment |
Is the new preprint “An Algorithm For Factoring Integers” by Yingpu Deng and Yanbin Pan worth reading? @Someone: the use of the additional modulo operation was certainly inspired by that use of the modulo within the AKS test; however, the current authors fail to justify why the polynomial coefficients might be interesting. The AKS test uses the fact that if the result of that is not congruent to the polynomial $x^n + a$, then $n$ is not prime; AKS says nothing about what the polynomial may be if it is not congruent, and the paper gives no justification (either theoretical or experimental) for us the expect the coefficients of that polynomial to be at all interesting. |
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Mar 1 |
answered | Is the new preprint “An Algorithm For Factoring Integers” by Yingpu Deng and Yanbin Pan worth reading? |
