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| visits | member for | 1 year, 8 months |
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| stats | profile views | 18 |
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May 17 |
revised |
Polynomial multiplication and division in 2^128 changed wording to say more clearly what the OP wants to do |
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May 17 |
suggested | suggested edit on Polynomial multiplication and division in 2^128 |
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May 12 |
comment |
Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? (continued from previous comment) an initial load $(a_2,a_1,a_0)$ and feedback taps $f_1,f_2,f_3$ where $$\begin{align}a_3&=a_2f_1+a_1f_2+a_0f_3\\a_4&=a_3f_1+a_2f_2+a_1f_3\\a_5&=a_4f_1+a_3f_2+a_2f_3\end{align}$$ so that we need $6$ bits of the sequence, not $3$, to get $3$ linear equations to solve for $f_1,f_2,f_3$. This is the same as the amount of information needed by the Berlekamp-Massey algorithm, but the Berlekamp-Massey algorithm will also find the shortest LFSR that generates any arbitrarily $6$-bit sequence, not just for the sequences known to be generated by a $n$-bit LFSR |
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May 12 |
comment |
Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? This answer is incorrect in the details. If one knows that the sequence is generated by an LFSR of length $128$ bits, then there are $128$ coefficients of the feedback polynomial (a.k.a. tap locations) that need to be determined, and $128$ bits of the sequence (exactly the initial loading of the LFSR) are not enough to determine these feedback coefficients; you need $256$ bits. Take a simpler case: for $n=3$, we have (continued in next comment) |
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May 11 |
comment |
Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? @geo909 OK, I am still a little confused. If $f(a,b)=a\vee b = a \oplus b \oplus ab$, does $f(L_1,L_2)$ equal $L_1+L_2+L_1L_2$ for relatively prime $L_1, L_2$? Or does it equal something else? |
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May 11 |
comment |
Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? @geo909 Something is awry in your comment. The combiner function $f$ is a Boolean function that maps $m$ bits (the LFSR outputs) to $1$ bit, the desired sequence with high linear complexity. So what does $f(L_1,L_2,\ldots, L_m)$ mean since the arguments and output have changed from bits to integers? |
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May 11 |
revised |
Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? added 1 characters in body |
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May 11 |
comment |
Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? In my previous comment, the last clause, which appears in parentheses, should read "and the initial loading is not all-zeroes" The word not was inadvertently left out, and it is now too late to edit the comment. |
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May 11 |
revised |
Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? added 475 characters in body |
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May 11 |
comment |
Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? The edit by rphv has many ambiguities in it. A linear feedback shift register (LFSR) of length $n$ does not necessarily generate a sequence of period $2^n-1$. The period could be much smaller and could depend on the initial loading and the feedback polynomial, that is, for some feedback polynomials, one can get sequences of different periods by changing the initial loading. It is true that the maximum period is $2^n-1$ and occurs when the feedback polynomial is a primitive polynomial in the sense that coding theorists use the term (and the initial loading is all-zeroes). |
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May 11 |
answered | Combining LFSRs for Stream Ciphers: Why do we need high non-linearity? |
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May 7 |
comment |
What does $(\mathbb{Z}_n^*)^2$ mean? @RickyDemer Thanks for catching that. Unfortunately, it is too late to edit my comment. |
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May 7 |
comment |
What does $(\mathbb{Z}_n^*)^2$ mean? Typically, $\mathbb Z_n^*$ would mean the multiplicative group of the ring $\mathbb Z_n$, the set of integers $\{0,1,2,\ldots, n-1\}$ with arithmetic being done modulo $n$. The multiplicative group would, of course, be $\{1,2,\ldots, n-1\}$ with multiplication being done modulo $n$. I would suspect that $(\mathbb{Z}_n^*)^2$ is the Cartesian product $\mathbb{Z}_n^*\times \mathbb{Z}_n^*$ whose elements are of the form $(a,b)$ with $a, b \in \mathbb{Z}_n^*$. |
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May 7 |
revised |
Implementing secret reconstruction in Shamir's Secret Sharing edited tags |
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May 7 |
answered | Implementing secret reconstruction in Shamir's Secret Sharing |
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Apr 25 |
revised |
repeating-key xor and hamming distance included clarification that the length of $X$ here is the length in bytes, not the length in bits. |
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Apr 25 |
suggested | suggested edit on repeating-key xor and hamming distance |
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Apr 25 |
comment |
repeating-key xor and hamming distance Could you cite where you read this claim? If the key $K$ is indeed of length $n$ bits and $X$ and $Y$ respectively bits $1$ through $n$ and bits $n+1$ through $2n$ of the plaintext, then the encrypted strings available to you are $X\oplus K$ and $Y\oplus K$, and the Hamming distance between them is the same as the Hamming distance between $X$ and $Y$. I don't see offhand why $X$ and $Y$ should be differing in only a few positions. |
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Mar 12 |
comment |
How to obtain a one-value share in Shamir's secret sharing Shamir's secret-sharing scheme does not need to have share #$3$ be the value of that polynomial $f(x)$ at $x=3$; the $n$ share-values are the values of $f(x)$ at $x_1, x_2, \ldots, x_n$ where the $x_i$ are $n$ distinct nonzero elements of the field. One does not have to choose $x_i = i$. But, of course, what each share holder is given is $(x_i, f(x_i))$, or, possibly, $(g(x_i), f(x_i))$ where $g(x_i)$ is the result of encrypting $x_i$ via an encryption scheme known to the trusted authority who will reconstruct the secret. |
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Mar 8 |
answered | Is there a way to use Shamir Secret Sharing with updatable data? |
