NoteNextra-origin/content/CSE5313/CSE5313_L10.md

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

Proof

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

So H G^T = 0.

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).

Proof

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.

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.

Proof

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.

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.

Example of Reed-Muller code

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}

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

Question 4

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

Proof

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.