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5

Multiplication of bits matrices works just like multiplication of number matrices, except the rule of addition is modified to: $1+1\mapsto 0$. Let $U$ (resp. $V$) be a square matrix of $n\times n$ elements noted $u_{l,c}$ (resp. $v_{l,c}$) with $1\le l\le n$ and $1\le c\le n$. The product $U\cdot V$ is a square matrix $W$ of $n\times n$ elements noted $w_{l,...


5

One observation is that if we modify the problem so that $M, A, B$ are random invertible matrices, then it is easy to prove the security of the system. In fact, we can prove that the system is informationally secure; that is, for any observed $C_1, C_2$ pair, for any possible value of $K$, there is a unique set of values of $A, B, M$ that yield that $K$ (...


3

We call an operation F linear if the following holds: $F(X+Y) = F(X) + F(Y)$ for all $X, Y$ within the appropriate set, and for some group operator $+$. Now, if we consider matrix multiplication by a fixed matrix $A$, we do have the identity: $A \cdot (X+Y) = A \cdot X + A \cdot Y$ for arbitrary vectors $X$, $Y$, and where $+$ is vector addition. Hence,...


3

The matrix is not MDS over $GF(2)$; No binary MDS codes exist and non nonbinary (over $GF(2^n)$ MDS codes would have this generator whose scalar entries are in the field $GF(2)$). Over $GF(2^n)$ The branch number, which is the minimum weight of the corresponding linear code is 4, in $GF(2^n)$ for all $n$. This covers all possible fields of interest for ...


2

Theorem Let $A:{GF(2^m)}^n \to {GF(2^m)}^n$ be an $n\times n \{0,1\}$-matrix over $GF(2^m)$. Then the branch number of $A$ is at most $\frac{2n+4}{3}$. Let $A$ be the almost-MDS matrix. So its branch number is $n$. Using above theorem we have $$n \leq \frac{2n+4}{3} $$ So $n=2,3$ or $4$. The size of matrix is a tool for finding upper bound of branch ...


2

Slide #8 in the presentation you linked to describes the way Käsper and Schwabe pack the bits of the AES data blocks into CPU registers. According to the slide, what they're doing is processing eight 128-bit AES blocks in parallel, using eight 128-bit XMM registers to store them. They're not doing basic "naïve bitslicing", which would involve using 128 $n$-...


2

You misunderstand $(02) \cdot 10000100$; it is not integer multiplication (resulting in a 9 bit integer); instead, it is multiplication in $GF(2^8)$ (which results in an element in $GF(2^8)$, which can be represented in 8 bits). AES uses a polynomial representation of $GF(2^8)$, using the polynomial $x^8 + x^4 + x^3 + x + 1$; what this means is that ...


2

You would need (at least) 3 pairs of vectors in order to determine the 3*3 matrix.


2

Let $k=\left(\begin{array}{cc}k_0 & k_1\\k_2 & k_3\end{array}\right)$ be the key. And I'll assume the transformation $a=0$, $b=1$, and so on. So you know $k\left(\begin{array}{c}0\\1\end{array}\right)=\left(\begin{array}{c}c_0\\c_1\end{array}\right)$ where you know $c_0$ and $c_1$. Thus $c_0=0k_0+1k_1$ and $c_1=0k_2+1k_3$. So that is two equations, ...


1

Assume that we have to compute $M\times x$, where $M$ is a $n\times n$ matrix, and $x$ is a $n\times 1$ vector, all entries of $M$ and $x$ are in $GF(2^8)$. We have: $$ M\times x = M \times \left( \begin{matrix} x_0 \\ x_1 \\ \vdots \\x_{n-1}\end{matrix}\right) = M \times \left( \begin{matrix} x_0 \\ 0\\ \vdots \\ 0\end{matrix}\right) \oplus M \times \...


1

In literature there are 2 ways to show the affine transform for a given polynomial, and that depends on the location of the MSB in the input as a polynomial. The polynomial representation of the full transformation, in the format of the original Rijndael paper, is: $b(x) = a(x)(x^7 + x^6 + x^5 + x^4 + 1) + (x^7 + x^6 + x^2 + x) ~~mod~~ x^8 + 1$ Where $a(x)...


1

Well, each party would know that the other party uses the same key, because they would probably have the same public key. Now, if some person I have never met before would use the same key as me, I would go into paranoid panic mode. That person could decrypt all messages that were only meant to be decrypted by me (and of course vice versa). However, if I ...


1

Both representations are essentially equivalent. If $$ \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} k_{11} & k_{12} & k_{13} \\ k_{21} & k_{22} & k_{23} \\ k_{31} & k_{32} & k_{33} \end{bmatrix} \cdot \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix}, $$ then, equivalently $$ \begin{bmatrix} p_1 & p_2 & ...



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