Question

Question 15: Let C be the BCH code of Examples 1 and 2, with codewords written as polynomials of degree . Suppose the codeword c (x)transmitted with errors in the coefficients of ${\mathit{x}}^{\mathbf{i}}$and ${\mathit{x}}^{\mathbf{j}}$ and role="math" localid="1659179965727" $\mathit{r}{\mathbf{\left(}\mathit{x}\mathbf{\right)}}^{}$ is received. Then $\mathit{D}\left(x\right)\mathbf{=}\left(x+{\alpha }^i\right)\left(x+{\alpha }^j\right)\mathbf{\in }\mathit{K}\left[x\right]$, whose roots are role="math" localid="1659179995612" ${\mathit{a}}^{\mathbf{i}}$and role="math" localid="1659179986054" ${\mathit{a}}^{\mathbf{j}}$ , is the error-locator polynomial. Express the coeffecients ofD(x)in terms of $\mathit{r}\left({\alpha }^{}\right)\mathbf{,}\mathit{r}\left({\alpha }^2\right)\mathbf{,}\mathit{r}\left({\alpha }^3\right)$as follows.

1. Show that $\mathit{r}\left(x\right)\mathbf{-}\mathit{c}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{i}}\mathbf{+}{\mathit{x}}^{\mathbf{j}}$.
2. Show that $\mathit{r}\left({\alpha }^k\right)\mathbf{=}{\mathit{\alpha }}^{\mathbf{k}\mathbf{i}}\mathbf{+}{\mathit{\alpha }}^{\mathbf{k}\mathbf{j}}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}$ .[ See the bold face statement on page 495.]
3. Show that $\mathit{D}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\left({\alpha }^i+{\alpha }^j\right)\mathit{x}\mathbf{+}{\mathit{\alpha }}^{\mathbf{i}\mathbf{+}\mathbf{j}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{r}\left(\alpha \right)\mathit{x}\mathbf{+}{\mathit{\alpha }}^{\mathbf{i}\mathbf{+}\mathbf{j}}$ .
4. Show that ${\mathit{\alpha }}^{\mathbf{i}\mathbf{+}\mathbf{j}}\mathbf{=}\mathit{r}\left({\alpha }^2\right)\mathbf{+}\frac{\mathbf{r}\left({\alpha }^3\right)}{\mathbf{r}\left(\alpha \right)}$ . [Hint: Show that $\mathit{r}{\left(\alpha \right)}^{\mathbf{3}}\mathbf{=}{\left({\alpha }^i+{\alpha }^j\right)}^{\mathbf{3}}\mathbf{=}{\mathit{\alpha }}^{\mathbf{3}\mathbf{i}}\mathbf{+}{\mathit{\alpha }}^{\mathbf{3}\mathbf{j}}\mathbf{+}{\mathit{\alpha }}^{\mathbf{i}\mathbf{+}\mathbf{j}}\left({\alpha }^i+{\alpha }^j\right)\mathbf{=}\mathit{r}\left({\alpha }^3\right)\mathbf{+}\mathit{r}\left(\alpha \right){\mathit{\alpha }}^{\mathbf{i}\mathbf{+}\mathbf{j}}$and solve for ,${\mathit{\alpha }}^{\mathbf{i}\mathbf{+}\mathbf{j}}$note that$\mathit{r}{\left(\alpha \right)}^{\mathbf{2}}\mathbf{=}\mathit{r}\left({\alpha }^2\right)$ .]

Short answer

Answer

$$\begin{gathered} \text{a.Ithasbeenprovedthat}r\left(x\right)-c\left(x\right)=x^i-x^j \\ \text{b.Ithasbeenprovedthat}r\left({\alpha }^k\right)={\alpha }^{ki}+{\alpha }^{kj} \\ \text{c.Ithasbeenprovedthat}D\left(x\right)=x^2+\left({\alpha }^i+{\alpha }^j\right)x+{\alpha }^{i+j}=x^2+r\left(\alpha \right)+{\alpha }^{i+j} \\ \text{d.Ithasbeenprovedthat}{\alpha }^{i+j}=r\left({\alpha }^2\right)+\frac{r\left({\alpha }^3\right)}{r\left(\alpha \right)} \end{gathered}$$

Step-by-step solution

Write the given data

Given that C are the BCH code of Examples 1 and 2, with codewords written as polynomials of degree .

The codeword transmitted with errors in the coefficients of xⁱ and x^j and r(x) is received.

$\mathbf{D}\left(\mathbf{x}\right)\mathbf{=}\left(\mathbf{x}\mathbf{+}{\mathbf{\alpha }}^{\mathbf{i}}\right)\left(\mathbf{x}\mathbf{+}{\mathbf{\alpha }}^{\mathbf{j}}\right)\in \mathbf{K}\left[\mathbf{x}\right]$, whose roots are aⁱ and a^j, is the error-locator polynomial.

rx-cx=xi+xjStep 2: Show that  (a)

Since, the codeword c (x) transmitted with errors in the coefficients of and and r (x) is received, therefore codeword c (x) and received word differ exactly by two places $x^i$ and $x^j$.

Hence $r\left(x\right)-c\left(x\right)=x^i-x^j$.

Show that  (b)

By the definition of g(x) we have $g\left(a^k\right)=0$ for $k=1,2,3,4$.

Since c(x) is a codeword it is a multiple of g(x).

Hence, from part (a) $r\left({\alpha }^k\right)={\alpha }^{ki}+{\alpha }^{kj}$.

Show that  D(x)=x2+(αi+αj)x+αi+j=x2+r(α)x+αi+j(c)

$$\begin{gathered} Byexample2D\left(x\right)=x^2+r\left(\alpha \right)x+{\alpha }^{i+j}. \\ Bypart\left(b\right),r\left({\alpha }^k\right)={\alpha }^{ki}+{\alpha }^{kj}. \\ ThusD\left(x\right)=x^2+\left({\alpha }^i+{\alpha }^j\right)x+{\alpha }^{i+j} \\ Hence,D\left(x\right)=x^2+\left({\alpha }^i+{\alpha }^j\right)x+{\alpha }^{i+j}=x^2+r\left(\alpha \right)+{\alpha }^{i+j} \end{gathered}$$

Show that  αi+j=r(α2)+r(α3)r(α)∅d

Now, write the equation as:

$$\begin{gathered} r{\left(\alpha \right)}^3={\left({\alpha }^i+{\alpha }^j\right)}^3 \\ ={\alpha }^{2i}+{\alpha }^{3j}+{\alpha }^{i+j}\left({\alpha }^i+{\alpha }^j\right) \\ =r\left({\alpha }^3\right)+{\alpha }^{i+j}r\left(\alpha \right) \end{gathered}$$

Therefore, ${\alpha }^{i+j}=r{\left(\alpha \right)}^2+\frac{r\left({\alpha }^3\right)}{r\left(\alpha \right)}$

Using Freshman’s Dream, $r{\left(\alpha \right)}^2=r\left({\alpha }^2\right)$.

Hence${\alpha }^{i+j}=r\left({\alpha }^2\right)+\frac{r\left({\alpha }^3\right)}{r\left(\alpha \right)}$