Posted on Categories:Cryptography, 密码学, 数学代写

# 数学代写|密码学代写Cryptography Theory代考|Mix Columns

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|密码学Cryptography Theory代考|Mix Columns

In this step, each column of the state is viewed as a polynomial of degree 3 or less. For example, the following column
$$\left(\begin{array}{l} a_0 \ a_1 \ a_2 \ a_3 \end{array}\right)$$
is viewed as $a(x)=a_3 x^3+a_2 x^2+a_1 x+a_0$. However, the coefficients, $a_3, a_2, a_1$, and $a_0$, are all bytes. That is, the coefficients themselves form polynomials that may be added or multiplied modulo the irreducible polynomial $x^8+x^4+x^3+x+1$ from the SubBytes step.

In the MixColumns step, each column, expressed as a polynomial, is multiplied by the polynomial $c(x)=3 x^3+x^2+x+2$. It is then reduced modulo $x^4+1$, so that it may still be expressed as a column (i.e., a polynomial of degree 3 or smaller).

Working modulo $x^4+1$ is a bit different than modulo $x^8+x^4+x^3+x+1$. First of all, $x^4+1$ is reducible! So a randomly chosen $c(x)$ needn’t be invertible. For this reason, $c(x)$ had to be chosen carefully, but how was $x^4+1$ chosen? It was picked so that products could be easily reduced. Moding out by $x^4+1$ is the same as defining $x^4=-1$, but $-1=1(\bmod 2)$, so we have $x^4=1$. This allows us to very easily reduce powers of $x$. We have $x^5=x, x^6=x^2, x^7=x^3$, and $x^8=x^0=1$. In general, $x^n=x^{n(\bmod 4)}$. Thus,
\begin{aligned} c(x) a(x)= & \left(3 x^3+x^2+x+2\right)\left(a_3 x^3+a_2 x^2+a_1 x+a_0\right) \ = & 3 a_3 x^6+3 a_2 x^5+3 a_1 x^4+3 a_0 x^3 \ & +a_3 x^5+a_2 x^4+a_1 x^3+a_0 x^2 \ & +a_3 x^4+a_2 x^3+a_1 x^2+a_0 x \ & +2 a_3 x^3+2 a_2 x^2+2 a_1 x+2 a_0 \end{aligned}

reduces to
\begin{aligned} c(x) a(x)= & 3 a_3 x^2+3 a_2 x+3 a_1+3 a_0 x^3 \ & +a_3 x+a_2+a_1 x^3+a_0 x^2 \ & +a_3+a_2 x^3+a_1 x^2+a_0 x \ & +2 a_3 x^3+2 a_2 x^2+2 a_1 x+2 a_0 . \end{aligned}

Finally, we involve the key! This is simply an XOR (self inverse) of each byte of the state with a byte of the key for the relevant round. Each round uses a distinct key derived from the original key. This is done as follows.

First, the original key is taken 32 bits at a time and placed at the beginning of what will become the “expanded key.” This expanded key will eventually be divided into equal size pieces to provide the round keys, in order. For AES-128, the original key will serve to initialize the expanded key blocks $k_0, k_1, k_2, k_3$. For AES-196, $k_4$ and $k_5$ will also be filled at this point; for AES-256, $k_6$ and $k_7$ will be filled. Then, more 32 bit blocks are defined recursively. The formulas for each of the three key sizes follow. They all involve a function, $f$, which will be detailed shortly.
For 128-bit keys:
\begin{aligned} & k_i=k_{i-4} \oplus k_{i-1}, \text { if } i \neq 0(\bmod 4) \ & k_i=k_{i-4} \oplus f\left(k_{i-1}\right), \text { if } i=0(\bmod 4) \end{aligned}
For 196-bit keys:
\begin{aligned} & k_i=k_{i-6} \oplus k_{i-1}, \text { if } i \neq 0(\bmod 6) \ & k_i=k_{i-6} \oplus f\left(k_{i-1}\right), \text { if } i=0(\bmod 6) \end{aligned}
where $f$ consists of a circular left shift of 1 byte for the input, followed by a substitution using Rijndael’s $S$-box, for each byte, and finally an XOR of this result with the appropriate round constant, $R C$ (to be discussed).

## 数学代写|密码学Cryptography Theory代考|Mix Columns

$$\left(\begin{array}{llll} a_0 & a_1 & a_2 & a_3 \end{array}\right)$$

$$c(x) a(x)=\left(3 x^3+x^2+x+2\right)\left(a_3 x^3+a_2 x^2+a_1 x+a_0\right)=3 a_3 x^6+3 a_2 x^5+3 a_1 x^4+3 a_0 x^3+a_3 x^5+a_2 x^4+a_1 x^3+a_0 x^2$$
$$c(x) a(x)=3 a_3 x^2+3 a_2 x+3 a_1+3 a_0 x^3 \quad+a_3 x+a_2+a_1 x^3+a_0 x^2+a_3+a_2 x^3+a_1 x^2+a_0 x \quad+2 a_3 x^3+2 a_2 x^2+2 a_1 x+2 a_0$$

$k_0, k_1, k_2, k_3$. 对于 AES-196， $k_4$ 和 $k_5$ 也将在这一点上被填补；对于 AES-256， $k_6$ 和 $k_7$ 会被填 满。然后，递归地定义更多的 32 位块。三个密钥大小中的每一个的公式如下。它们都涉及一个功 能， $f$ ，稍后会详细介绍。

$k_i=k_{i-4} \oplus k_{i-1}$, if $i \neq 0(\bmod 4) \quad k_i=k_{i-4} \oplus f\left(k_{i-1}\right)$, if $i=0(\bmod 4)$

$k_i=k_{i-6} \oplus k_{i-1}$, if $i \neq 0(\bmod 6) \quad k_i=k_{i-6} \oplus f\left(k_{i-1}\right)$, if $i=0(\bmod 6)$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。