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## 数学代写|编码理论代写Coding theory代考|DETERMINING MINIMAL POLYNOMIALS

If $w$ is a root in $\mathrm{GF}\left(q^{m}\right)$ of an irreducible polynomial of degree $m$ over $\mathrm{GF}(q)$, then every element in $\mathrm{GF}\left(q^{m}\right)$ may be expressed as a polynomial over $\mathrm{GF}(q)$ of degree $<m$ in $w$. Each element in $\mathrm{GF}\left(q^{m}\right)$ is a root of some minimal polynomial over $\mathrm{GF}(q)$, and the degree of this minimal polynomial is a divisor of $m$. Often one wishes to determine this minimal polynomial.

For example, suppose $w$ is a root in $\mathrm{GF}\left(2^{6}\right)$ of the cyclotomic polynomial $Q^{(9)}(x)=x^{6}+x^{3}+1$. Since the multiplicative order of $2 \bmod 9$ is $6, w$ has 6 distinct conjugates in $\mathrm{GF}\left(2^{6}\right)$, and $Q^{(9)}(x)$ is irreducible over $\mathrm{GF}(2)$. Suppose $u=w^{3}+w+1$. What is the minimal polynomial of $u$ ?

The most straightforward approach is to express $u, u^{2}, u^{3}, \ldots, u^{6}$ as polynomials in $w$ and to determine which linear combination of these is 0. In the present example, we have
\begin{aligned} 1 &=1 & & \ u &=1+w & &+w^{3} \ u^{2} &=& w^{2}+w^{3} \ u^{3} &=1 & &+w^{2}+w^{3}+w^{4}+w^{5} \ u^{4} &=1 & &+w^{3}+w^{4} \ u^{5} &=& & w^{3}+w^{4}+w^{5} \ u^{6} &=& & w+w^{2}+w^{3}+w^{4}+w^{5} \end{aligned}
Solving the equations
$$\left[M_{0}, M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6}\right]\left[\begin{array}{llllll} 1 & 0 & 0 & 0 & 0 & 0 \ 1 & 1 & 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 & 0 \ 1 & 0 & 1 & 1 & 1 & 1 \ 1 & 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 1 & 1 & 1 \ 0 & 1 & 1 & 1 & 1 & 1 \end{array}\right]=[0,0,0,0,0,0]$$
we obtain $\mathbf{M}=[1,1,1,0,0,1,1]$, so the minimal polynomial of $u$ is $1+x+$ $x^{2}+x^{5}+x^{6}$.

This method has the advantage of great generality. It is also easily programmed on a computer, and it is quite feasible even for large fields. In certain special cases, however, hand computations may be considerably shortened by the use of appropriate shortcuts which we now present.

## 数学代写|编码理论代写Coding theory代考|REORDERING THE COLUMNS OF THE PARITY-CHECK MATRIX OF HAMMING CODES

In Sec. 1.3, we saw how to construct single-error-correcting codes having $n=2^{m}-1$ digits, of which $m$ are check digits and $n-m$ are information digits. Each of the $n$ columns of the parity-check matrix must contain a different nonzero binary $m$-tuple, which is called the location number of that digit. As long as the $n$ different digits of the code are assigned different nonzero location numbers, it does not matter how these location numbers are ordered. For example, the parity-check matrix for a single-error-correcting code of block length 15 might be given as
$$\mathcal{F}^{\prime}=\left[\begin{array}{lllllllllllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \end{array}\right]$$
In order to correct additional errors, we must add more rows to this matrix. In order to find intelligent methods for adding those additional rows and in order to decode the resulting codes, we found it helpful to consider each of these $n$ error location numbers as a nonzero element in $\mathrm{GF}\left(2^{m}\right)$, represented as a binary polynomial of degree $<m$ in a root of some irreducible binary polynomial of degree $m$. We shall now show that this viewpoint is also helpful for designing efficient encoders and decoders, even in the single-error-correcting case.

In the present example, with $m=4$, we might choose the primitive polynomial to be $x^{4}+x+1$. The leftmost column then has location number 1 ; the second column, location number $\alpha$; the third column, location number $\alpha+1 ; \ldots$; the eleventh column, $\alpha^{3}+\alpha+1 ; \ldots .$ If we denote the error location numbers by the field elements $\eta_{i}^{\prime}, i=0,1$, $2, \ldots, 14$, we have $\eta_{0}^{\prime}=1, \eta_{1}^{\prime}=\alpha, \eta_{2}^{\prime}=\alpha+1, \eta_{3}^{\prime}=\alpha^{2}, \eta_{4}^{\prime}=\alpha^{2}+1$, $\eta_{5}^{\prime}=\alpha^{2}+\alpha, \ldots, \eta_{14}^{\prime}=\alpha^{3}+\alpha^{2}+\alpha+1$. The 15-dimensional binary vector $\mathrm{R}=\left[R_{0}, R_{1}, \ldots, R_{14}\right]$ then has the syndrome $\exists \mathrm{C}^{\prime} \mathrm{R}^{t}$, which is the Galois field representation of the sum of the corresponding location numbers, $\xi C^{\prime} \mathbf{R}^{t}=\sum_{i=0}^{14} R_{i \eta_{i}^{\prime}}$.

## 数学代写|编码理论代写Coding theory代考|REORDERING THE COLUMNS OF THE PARITY-CHECK MATRIX

The syndrome, $\mathbf{S}^{t}=\xi c R^{t}$, consists of $2 m$ digits. The first $m$ digits of the syndrome give $S_{1}$, the sum of the error locetions; the second $m$ digits of the syndrome give $S_{3}$, the sum of the cubes of the error locations. If the received vector is represented by the polynomial $R(x)=\sum_{i=0}^{n-1} R_{i} x^{i}$, then $S_{1}-R(u)$, and $S_{3}-R\left(\alpha^{3}\right)$. These two power sums may be computed from the received word separately. To compute $S_{1}$, we divide $R(x)$ by $M^{(1)}(x)$, the minimal polynomial of $\alpha$, to obtain the remainder $r^{(1)}(x)$. We then have $S_{1}=r^{(1)}(\alpha)$. To compute $S_{3}$, we divide $R(x)$ by

$M^{(3)}(x)$, the minimal polynomial of $\alpha^{3}$, to obtain the remainder $r^{(3)}(x)$. We then compute $S_{3}=r^{(3)}\left(\alpha^{3}\right)$. In general, the calculation of $S_{3}$ from $r^{(3)}$ may require as many as $m$ parity checks, and each of these parity checks has at most $m$ inputs. $\dagger$

In the present example, the minimal polynomial of $\alpha^{3}$ is $x^{4}+x^{3}+$ $x^{2}+x+1$. The polynomials $r^{(1)}(x)$ and $r^{(3)}(x)$ may be obtained by running the received word through the horizontal feedback shift registers shown in Fig. 5.06. As the decoder receives the digits from the channel, they are fed into the 15-digit received-word buffer, shown horizontally at the top of Fig. 5.06. After all 15 digits of the block are received, this buffer contains the polynomial $R(x)$, representing the received word. As the received channel digits are fed into this register, they are simultaneously fed into the two feedback shift registers, wired according to the minimal polynomials of $\alpha$ and $\alpha^{3}$. These registers are initially zero. As the digits arrive from the channel, these registers divide the polynomial $R(x)$ by $M^{(1)}$ and $M^{(3)}$, the minimal polynomials of $\alpha$ and $\alpha^{3}$, respectively. When the entire block of 15 digits has arrived, these registers contain the remainders $r^{(1)}(x)$ and $r^{(3)}(x)$.

## 数学代写|编码理论代写Coding theory代考|DETERMINING MINIMAL POLYNOMIALS

$$1=1 \quad u \quad=1+w \quad+w^{3} u^{2}=w^{2}+w^{3} u^{3}=1 \quad+w^{2}+w^{3}+w^{4}+w^{5} u^{4} \quad=1 \quad+w^{3}+w^{4} u^{5}=\quad w^{3}+w^{4}+$$

## 数学代写|编码理论代写Coding theory代考|REORDERING THE COLUMNS OF THE PARITY-CHECK MATRIX

$M^{(3)}(x)$ ，的最小多项式 $\alpha^{3}$ ，得到余数 $r^{(3)}(x)$. 然后我们计算 $S_{3}=r^{(3)}\left(\alpha^{3}\right)$. 一般来说，计算 $S_{3}$ 从 $r^{(3)}$ 可能需要多达 $m$ 奇偶校验，每个奇偶 校验最多有 $m$ 输入。 $\dagger$

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## 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。

assignmentutor™您的专属作业导师
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