# 数学代写|编码理论代写Coding theory代考|ELEC7604

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## 数学代写|编码理论代写Coding theory代考|THE KEY EQUATION FOR DECODING BINARY BCH CODES

If the encoder transmits the binary $\mathrm{BCH}$ codeword
$$C(x)=\sum_{i=0}^{n-1} C_{i} x^{i}$$
and the channel noise causes additive errors given by the coefficients of the binary polynomial
$$E(x)=\sum_{i=0}^{n-1} E_{i} x^{i}$$
then the received word will be given by
$$R(x)=\sum_{i=0}^{n-1} R_{i} x^{i}=\sum_{i=0}^{n-1} C_{i} x^{i}+\sum_{i=0}^{n-1} E_{i} x^{i}$$
For $j=1,2, \ldots, 2 t$, the codeword is a multiple of the minimal polynomial of $\alpha^{j}$, and therefore
$$R\left(\alpha^{j}\right)=0+\sum_{i=0}^{n-1} E_{i} \alpha^{j i}=\sum_{k=1}^{e} X_{k}^{j}=S_{j}$$
where we let the Galois field error locations $X_{1}, X_{2}, \ldots, X_{e}$ denote the positions where $E_{i}=1$. If $R(x)$ is divided by $M^{(j)}(x)$, the minimal polynomial of $\alpha^{j}(1 \leq j \leq 2 t)$, then $S_{j}=R\left(\alpha^{j}\right)$ may be readily computed from the remainder, $r^{(j)}\left(\alpha^{j}\right)$, as in Sec. 5.2. After the decoder has calculated $S_{1}, S_{2}, S_{3}, \ldots, S_{2 t}$, the major problem is to find the error locations, $X_{1}, X_{2}, \ldots, X_{e}$, from the equations
$$\sum_{i=1}^{e} X_{i}^{j}=S_{j} \quad j=1,2,3, \ldots, 2 t$$
In general, these equations will have many solutions, each corresponding to a different error pattern in the same coset of the additive group of codewords. The decoder must find a solution with as small a value of $e$ as possible.

In order to solve these equations, the decoder first attempts to find the coefficients of the error-locator polynomial, defined by Definition $7.21$ [or Eq. (1.46)].

## 数学代写|编码理论代写Coding theory代考|HEURISTIC SOLUTION OF THE KEY EQUATION

We wish to solve the key equation
$$(1+S) \sigma \equiv \omega \bmod z^{2 t+1}$$
for the polynomials $\sigma(z)$ and $\omega(z)$, given $S(z) \bmod z^{2 t+1}$. The problem looks difficult, so we break it up into smaller pieces. We consider the sequence of equations
$$(1+S) \sigma^{(k)} \equiv \omega^{(k)} \bmod z^{k+1}$$
For each $k=0,1,2, \ldots, 2 t$, we shall find polynomials
$$\sigma^{(k)}=\sum_{i} \sigma_{i}^{(k)} z^{i}$$

and
$$\omega^{(k)}=\sum_{i} \omega_{i}^{(k)} z^{i}$$
which solve this equation. In general, these equations may have many solutions. Since the degree of $\sigma$ is the number of errors, a good decoder must attempt to find a solution in which degree $\sigma$ and degree $\omega$ are “small.”

If we have a solution to $(7.301)$, then we can hope that this same pair of polynomials, $\sigma^{(k)}$ and $\omega^{(k)}$, might also solve the equation
$$(1+S) \sigma^{(k)} \stackrel{?}{\equiv} \omega^{(k)} \bmod z^{k+2}$$
In general, we cannot expect to be so lucky. However, we may write
$$(1+S) \sigma^{(k)} \equiv \omega^{(k)}+\Delta_{1}^{(k)} z^{k+1} \bmod z^{k+2}$$

## 数学代写|编码理论代写Coding theory代考|SIMPLIFICATIONS IN ALGORITHM 7.4 FOR BINARY BCH CODES

Given any sequence $S_{1}, S_{2}, S_{3}, \ldots, S_{2 t}$, Algorithm $7.4$ will find lowdegree polynomials $\sigma(z)$ and $\omega(z)$ which solve the key equation (7.23). In the case of binary $\mathrm{BCH}$ codes, however, the sequence $S_{1}, S_{2}, S_{3}, \ldots$, $S_{2 t}$ is not arbitrary. The $S^{\prime}$ s must be power-sum symmetric functions of $e$ error locations, $X_{1}, X_{2}, \ldots, X_{e}$. Even if $e>t$, we have $S_{k}=$ $\sum_{i=1}^{e} X_{i}^{k}, S_{2 k}=\sum_{i=1}^{e} X_{i}{ }^{2 k}=\left(\sum_{i=1}^{e} X_{i}^{k}\right)^{2}=S_{k}{ }^{2}$. Thus, the $S^{\prime}$ s are related to each other by the equations $S_{2 k}=S_{k}{ }^{2}$. The generating function $S(z)$ must therefore satisfy the equation $[S(z)]^{2}=\sum_{k=1}^{\infty} S_{2 k} z^{2 k}=\widehat{S}(z)$, where I again use the accent ${ }^{\wedge}$ to denote the even part and the accent $\sim$ to denote the odd part, as in Sec. 3.2. The generating-function equation $\dot{S}=S^{2}$ leads to a considerable simplification of the iterative algorithm, as we shall now show.

Lemma 7.61 Let $1+S$ and $1+R$ be reciprocal generating functions in a field of characteristic $2:(1+S)(1+R)=1 .$ Then $\hat{R}=0$ iff $\hat{S}=S^{2}$

Proof Separating $(1+S)(1+R)=1$ into even and odd parts gives $(1+\hat{S})(1+\hat{R})+\widehat{S} \tilde{R}=1$ and $\widehat{S}(1+\hat{R})+(1+\hat{S}) \widetilde{R}=0$. Subtracting $\overleftrightarrow{S}$ times the latter from $1+\widehat{S}$ times the former gives $\left(1+\hat{S})^{2}-\widehat{S}^{2}\right=1+\hat{S}$, from which
$$\tilde{R}=\frac{1+\hat{S}}{(1+\hat{S})^{2}-\bar{S}^{2}}-1$$
In a field of characteristic $2,(1+\hat{S})^{2}-\widehat{S}^{2}=1+\widehat{S}^{2}+\widehat{S}^{2}=1+$ $(\hat{S}+\bar{S})^{2}=1+S^{2}$ and $\hat{R}=0$ iff $1+\hat{S}=1+S^{2}$ or $\hat{S}=S^{2}$
Q.E.D.

## 数学代写|编码理论代写Coding theory代考|THE KEY EQUATION FOR DECODING BINARY BCH CODES

$$C(x)=\sum_{i=0}^{n-1} C_{i} x^{i}$$

$$E(x)=\sum_{i=0}^{n-1} E_{i} x^{i}$$

$$R(x)=\sum_{i=0}^{n-1} R_{i} x^{i}=\sum_{i=0}^{n-1} C_{i} x^{i}+\sum_{i=0}^{n-1} E_{i} x^{i}$$

$$R\left(\alpha^{j}\right)=0+\sum_{i=0}^{n-1} E_{i} \alpha^{j i}=\sum_{k=1}^{e} X_{k}^{j}=S_{j}$$

$$\sum_{i=1}^{e} X_{i}^{j}=S_{j} \quad j=1,2,3, \ldots, 2 t$$

## 数学代写|编码理论代写Coding theory代考|HEURISTIC SOLUTION OF THE KEY EQUATION

$$(1+S) \sigma \equiv \omega \bmod z^{2 t+1}$$

$$(1+S) \sigma^{(k)} \equiv \omega^{(k)} \bmod z^{k+1}$$

$$\sigma^{(k)}=\sum_{i} \sigma_{i}^{(k)} z^{i}$$

$$\omega^{(k)}=\sum_{i} \omega_{i}^{(k)} z^{i}$$

$$(1+S) \sigma^{(k)} \stackrel{?}{\equiv} \omega^{(k)} \bmod z^{k+2}$$

$$(1+S) \sigma^{(k)} \equiv \omega^{(k)}+\Delta_{1}^{(k)} z^{k+1} \bmod z^{k+2}$$

QED

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