### 数学代写|编码理论代写Coding theory代考|MTH 4107

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• Statistical Computing 统计计算
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|编码理论代写Coding theory代考|MULTIPLICATIVE INVERSION

Let us now consider the problem of finding the multiplicative inverse of an element in the field of residue classes mod an irreducible binary polynomial $M(x)$ of degree $m$. Given the residue class containing $r(x)$, a polynomial of degree $<m$, we wish to find the polynomial $p(x)$ of degree $<m$ such that the product satisfies
$$r(x) p(x) \equiv 1 \bmod M(x)$$
or equivalently, $r(x) p(x)+M(x) q(x)=1$ for some polynomial $q(x)$. Since. $M(x)$ is irreducible, the ged of $M$ and $r$ is 1 . We may therefore apply the continued-fractions version of Euclid’s algorithm as described in Sec. 2.1. Starting with $r^{(-2)} \equiv M, r^{(-1)} \equiv r, p^{(-2)} \equiv 0, p^{(-1)} \equiv 1$, $q^{(-2)}=1, q^{(-1)}=0$, we use the division algorithm to find $a^{(k)}$ and $r^{(k)}$ such that
$$r^{(k-2)}=a^{(k)} r^{(k-1)}+r^{(k)} \quad \operatorname{deg} r^{(k)}<\operatorname{deg} r^{(k-1)}$$
We then set
\begin{aligned} &q^{(k)}=a^{(k)} q^{(k-1)}+q^{(k-2)} \ &p^{(k)}=a^{(k)} p^{(k-1)}+p^{(k-2)} \end{aligned}

The iteration is to be continued until $r^{(n)}=0$. The solution is then given by $q=q^{(n-1)}, p=p^{(n-1)}$ with $\operatorname{deg} q<\operatorname{deg} r, \operatorname{deg} p<\operatorname{deg} M=$ $m$. Since we wish to find only $p$ (and do not particularly care about $q$ ), we may dispense with the $q$ ‘s entirely.

Before designing the logical circuits, let us work an example. Suppose $r(x)=x^{4}+x+1$ and $M(x)=x^{5}+x^{2}+1$. One method of computing successive $a$ ‘s and $r$ ‘s and $p^{\prime}$ ‘s follows.

## 数学代写|编码理论代写Coding theory代考|MULTIPLICATION

When considering the multiplication of residue classes mod $M(x)$, where $M(x)$ is an irreducible binary polynomial of degree $m$, it is helpful to introduce the symbol $\alpha$ to denote the residue class containing $x$. Then $\alpha^{2}$ represents the residue class containing $x^{2}$, and, in general, if $r(x)$ is any polynomial, then $r(\alpha)$ represents the residue class containing $r(x)$. Since $M(x) \equiv 0 \bmod M(x)$, we must have $M(\alpha)=0$. The element represented by the symbol $\alpha$ is therefore a root of the polynomial $M(x)$. Hence, we have an obvious isomorphism between the field containing the $2^{m}$ residue classes $\bmod M(x)$ and the field containing the binary field and all polynomials in $\alpha$, where $\alpha$ is a root of the irreducible binary polynomial $M(x)$.

Any element $Y$ in this field may be expressed uniquely as a polynomial of degree $<m$ in $\alpha, Y=\sum_{i=0}^{m-1} Y_{i} \alpha^{i}$, where the $Y_{i}$ are binary numbers. The element $Y$ may be conveniently stored in an $m$-bit register, whose components contain the binary numbers $Y_{m-1}, Y_{m-2}, \ldots, Y_{0}$.

## 数学代写|编码理论代写Coding theory代考|MULTIPLICATION OF A REGISTER BY A WIRED CONSTANT

Let us first consider the multiplication of the field element in the $Y$ register by a constant field element $A$. We may assume that $A$ is represented by some binary polynomial in $\alpha$. Since $Y=\sum_{i=0}^{m-1} Y_{i} \alpha^{i}$, we have $Y A=\sum_{i=0}^{m-1} Y_{i}\left(A \alpha^{i}\right)$. Expressing $A \alpha^{i}$ as a polynomial of degree $<m$ in $\alpha$ gives $A \alpha^{i}=\sum_{j=0}^{m-1} A_{i, j} \alpha^{j}$, so that
\begin{aligned} Y A &=\sum_{i=0}^{m-1} Y_{i} \sum_{j=0}^{m-1} A_{i, j} \alpha^{j} \ &=\sum_{j=0}^{m-1}\left(\sum_{i=0}^{m-1} Y_{i} A_{i, j}\right) \alpha^{j} \end{aligned}
Thus, multiplication of the field element $Y$ by the field element $A$ is equivalent to multiplication of the $m$-dimensional binary row vector $\mathbf{Y}=\left[Y_{m-1}, Y_{m-2}, \ldots, Y_{0}\right]$ by the $m \times m$ matrix whose components are $A_{i, j}$. The rows of this matrix represent the products $A \alpha^{m-1}, A \alpha^{m-2}$, $\cdots, A$.

For example, let $M(x)=x^{5}+x^{2}+1$. Suppose we wish to multiply the contents of the $Y$ register by the field element $A=\alpha^{3}+\alpha$. We first compute
\begin{aligned} A \alpha &=\alpha^{4}+\alpha^{2} \ A \alpha^{2} &=\alpha^{5}+\alpha^{3}=\alpha^{3}+\alpha^{2}+1 \ A \alpha^{3} &=\alpha^{4}+\alpha^{3}+\alpha \ A \alpha^{4} &=\alpha^{5}+\alpha^{4}+\alpha^{2}=\alpha^{4}+1 \end{aligned}
The multiplication $Z=Y A$ is equivalent to
$$\left[Z_{4}, Z_{3}, Z_{2}, Z_{1}, Z_{0}\right]=\left[Y_{4}, Y_{3}, Y_{2}, Y_{1}, Y_{0}\right]\left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 1 \ 1 & 1 & 0 & 1 & 0 \ 0 & 1 & 1 & 0 & 1 \ 1 & 0 & 1 & 0 & 0 \ 0 & 1 & 0 & 1 & 0 \end{array}\right]$$
This multiplication may readily be accomplished by the circuit of Fig. 2.11.

## 数学代写|编码理论代写Coding theory代考|MULTIPLICATIVE INVERSION

$$r(x) p(x) \equiv 1 \bmod M(x)$$

$r^{(-2)} \equiv M, r^{(-1)} \equiv r, p^{(-2)} \equiv 0, p^{(-1)} \equiv 1, q^{(-2)}=1, q^{(-1)}=0$ ，我们使用除法算法找到 $a^{(k)}$ 和 $r^{(k)}$ 这样
$$r^{(k-2)}=a^{(k)} r^{(k-1)}+r^{(k)} \quad \operatorname{deg} r^{(k)}<\operatorname{deg} r^{(k-1)}$$

$$q^{(k)}=a^{(k)} q^{(k-1)}+q^{(k-2)} \quad p^{(k)}=a^{(k)} p^{(k-1)}+p^{(k-2)}$$

## 数学代写|编码理论代写Coding theory代考|MULTIPLICATION OF A REGISTER BY A WIRED CONSTANT

$$Y A=\sum_{i=0}^{m-1} Y_{i} \sum_{j=0}^{m-1} A_{i, j} \alpha^{j} \quad=\sum_{j=0}^{m-1}\left(\sum_{i=0}^{m-1} Y_{i} A_{i, j}\right) \alpha^{j}$$

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## MATLAB代写

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