# CSE5313 Coding and information theory for data science (Recitation 10)

## Question 2

Let $C$ be a Reed-Solomon code generated by

$$
G=\begin{bmatrix}
1 & 1 & \cdots & 1\\
\alpha_1 & \alpha_2 & \cdots & \alpha_n\\
\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\
\vdots & \vdots & \cdots & \vdots\\
\alpha_1^{k-1} & \alpha_2^{k-1} & \cdots & \alpha_n^{k-1}
\end{bmatrix}
$$

prove that there exists $v_1,v_2,\ldots,v_n\in \mathbb{F}_q\setminus \{0\}$ such that the parity check matrix is

$$
H=\begin{bmatrix}
1 & 1 & \cdots & 1\\
\alpha_1 & \alpha_2 & \cdots & \alpha_n\\
\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\
\vdots & \vdots & \cdots & \vdots\\
\alpha_1^{k-1} & \alpha_2^{k-1} & \cdots & \alpha_n^{k-1}
\end{bmatrix}\begin{bmatrix}
v_1 & 0 & \cdots & 0\\
0 & v_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & v_n
\end{bmatrix}
$$

### Some lemmas for linear codes

First we introduce the following lemmas for linear codes

Let $G$ and $H$ be the generator and parity-check matrices of (any) linear code

#### Lemma 1

$$
H G^T = 0
$$

<details>
<summary>Proof</summary>

By definition of generator matrix and parity-check matrix, $forall e_i\in H$, $e_iG^T=0$.

So $H G^T = 0$.
</details>

#### Lemma 2

Any matrix $M\in \mathbb{F}_q^{(n-k)\times n}$ such that $\operatorname{rank}(M) = n - k$ and $M G^T = 0$ is a parity-check matrix for $C$ (i.e. $C = \ker M$).

<details>
<summary>Proof</summary>

It is sufficient to show that the two statements

1. $\forall c\in C, c=uG, u\in \mathbb{F}^k$

$M c^T = M(uG)^T = M(G^T u^T) = 0$ since $M G^T = 0$.

Thus $C \subseteq \ker M$.

2. $\dim (\ker M) +\operatorname{rank}(M) = n$

We proceed by showing that $\dim (\ker M) =\dim (C)$.

Suppose $C$ does not span $\ker M$. Let $u_1,...,u_k$ be a basis for $C$. Then there exists $v\in \ker M\setminus C$.

By linear independence, if we have scalar $a_1,...,a_k$ and $b$ such that $a_1u_1+...+a_ku_k+bv=0$, then $a_1=...=a_k=b=0$. So $v=a_1u_1+...+a_ku_k$ for some $a_1,...,a_k$.

By definition of linear code, we have $v\in C$, contradicting the assumption.

</details>

### Solution

We proceed by applying the lemma 2.

1. $\operatorname{rank}(H) = n - k$ since $H$ is a Vandermonde matrix times a diagonal matrix with no zero entries, so $H$ is invertible.

2. $H G^T = 0$.

note that $\forall$ row $i$ of $H$, $0\leq i\leq n-k-1$, $\forall$ column $j$ of $G^T$, $0\leq j\leq k-1$

So

$$
\begin{aligned}
H G^T &= \begin{bmatrix}
1 & 1 & \cdots & 1\\
\alpha_1 & \alpha_2 & \cdots & \alpha_n\\
\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\
\vdots & \vdots & \cdots & \vdots\\
\alpha_1^{k-1} & \alpha_2^{k-1} & \cdots & \alpha_n^{k-1}
\end{bmatrix}\begin{bmatrix}
v_1 & 0 & \cdots & 0\\
0 & v_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & v_n
\end{bmatrix}\begin{bmatrix}
1 & \alpha_1 & \alpha_1^2 & \cdots & \alpha_1^{k-1}\\
1 & \alpha_2 & \alpha_2^2 & \cdots & \alpha_2^{k-1}\\
\vdots & \vdots & \vdots & \cdots & \vdots\\
1 & \alpha_n & \alpha_n^2 & \cdots & \alpha_n^{k-1}
\end{bmatrix}\\
&=
\begin{bmatrix}
1 & 1 & \cdots & 1\\
\alpha_1 & \alpha_2 & \cdots & \alpha_n\\
\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\
\vdots & \vdots & \cdots & \vdots\\
\alpha_1^{k-1} & \alpha_2^{k-1} & \cdots & \alpha_n^{k-1}
\end{bmatrix}
\begin{bmatrix}
v_1\\
v_2\\
\vdots\\
v_n
\end{bmatrix}\\
&=\sum_{l=1}^n\alpha_l^{r}v_l=0
\end{aligned}
$$

## Question 3

Show that in an MDS code $[n,k,d]_{\mathbb{F}_q}$, $d=n-k+1$, every $k$ entries determine the remaining $n-k$ entries of $G$.

That is, every $k\times k$ submatrix of $G$ is invertible.

<details>
<summary>Proof</summary>

Let $G$ be the generator matrix, and $G'$ be any $k\times k$ submatrix of $G$.

$$
G=\begin{bmatrix}
G'\in \mathbb{F}^{k\times k}|G''\in \mathbb{F}^{k\times (n-k)}
\end{bmatrix}
$$

We proceed by contradiction, suppose $G'$ is not invertible.

Then there exists $m',m''\in \mathbb{F}_q^k$ such that $m'\neq m''$ but $m'G'=m''G'$.

Note that $m'G=[m'G'|m'G'']$ and $m''G=[m''G'|m''G'']$.

So there are only $n-k$ entries of $m'G$ and $m''G$ are different, so $d\leq n-k$.

That violate with the assumption that $d=n-k+1$.
</details>

## Reed-Muller code

### Definition of Reed-Muller code (binary case)

$$
RM(r,m)=\left\{(f(\alpha_1),\ldots,f(\alpha_2^m))|\alpha_i\in \mathbb{F}_2^m,\deg f\leq r\right\}
$$

Length of $RM(r,m)$ is $2^m$.

<details>
<summary>Example of Reed-Muller code</summary>

Let $r=2$, $m=3$.

$\alpha_1=(0,0,0)$, $\alpha_2=(0,0,1)$, $\alpha_3=(0,1,0)$, $\alpha_4=(0,1,1)$, $\alpha_5=(1,0,0)$, $\alpha_6=(1,0,1)$, $\alpha_7=(1,1,0)$, $\alpha_8=(1,1,1)$.

$p(x)\deg\leq 2=\{1,x_1,x_2,x_3,x_1x_2,x_1x_3,x_2x_3\}$.

So $p(x)=1+x_1+x_2+x_3+x_1x_2+x_1x_3+x_2x_3$.

The generator matrix is defined by 

$$
G=\begin{bmatrix}1\\x_1\\x_2\\x_3\\x_1x_2\\x_1x_3\\x_2x_3\end{bmatrix}\begin{bmatrix}
1& 1&1&1&1&1&1&1\\
0& 0&0&0&1&1&1&1\\
0& 0&1&1&0&0&1&1\\
0& 1&0&1&0&1&0&1\\
0& 0&0&0&0&0&1&1\\
0& 0&0&0&0&1&0&1\\
0& 0&0&1&0&0&0&1\\
\end{bmatrix}
$$

</details>

So $\dim RM(r,m)=\sum_{i=0}^{r}\binom{m}{i}$.

## Question 4

$RM(m-1,m)$ is the parity check code.

<details>
<summary>Proof</summary>

By previous lemma, it is sufficient to show that $\dim (RM(m-1,m))=n-1$ and $RM(m-1,m)\subseteq \text{ parity code}$.

For the first property,

$$
\dim (RM(m-1,m))=\sum_{i=0}^{m-1}\binom{m}{i}=\sum_{i=0}^{m}\binom{m}{i}-\binom{m}{m}=2^m-1=n-1
$$

For the second property,

recall that $c=(c_1,c_2,\ldots,c_n)\in$ parity if $\sum_{i=1}^n c_i=0$.

So we need to show that $\sum_{i=1}^{2^m}f(\alpha_i)=0$ for every $f$ with $\deg f\leq m-1$.

Note that $\forall f\in RM(m-1,m)$, we can write $f(x_1,x_2,\ldots,x_m)=\sum_{S\subseteq [m],|S|\leq m-1}f_Sx_S$

So $\sum_{i=1}^{2^m}f(\alpha_i)=\sum_{S\subseteq [m],|S|\leq m-1}f_S\sum_{i=1}^{2^m}x_S(\alpha_i)=0$.

We denote $J(s)=\sum_{i=1}^{2^m}x_S(\alpha_i)$.

Note that $J(S)=\begin{cases}
1 & \text{if number of 1 in $S$ is odd}
0 & \text{otherwise}
\end{cases}$

And number of 1 in $S$ is $2^{m-|S|}$ which is even.

So $J(S)=0$ for all $S\subseteq [m],|S|\leq m-1$.
</details>