# CSE5313 Coding and information theory for data science (Lecture 7) ## Continue on Linear codes Let $\mathcal{C}= [n,k,d]_{\mathbb{F}}$ be a linear code. There are two equivalent ways to describe a linear code: 1. A generator matrix $G\in \mathbb{F}^{k\times n}_q$ with $k$ rows and $n$ columns, entry taken from $\mathbb{F}_q$. $\mathcal{C}=\{xG|x\in \mathbb{F}^k\}$ 2. A parity check matrix $H\in \mathbb{F}^{(n-k)\times n}_q$ with $(n-k)$ rows and $n$ columns, entry taken from $\mathbb{F}_q$. $\mathcal{C}=\{c\in \mathbb{F}^n:Hc^\top=0\}$ ### Dual code #### Definition of dual code $C^{\perp}$ is the set of all vectors in $\mathbb{F}^n$ that are orthogonal to every vector in $C$. $$ C^{\perp}=\{x\in \mathbb{F}^n:x\cdot c=0\text{ for all }c\in C\} $$ Also, the alternative definition is: 1. $C^{\perp}=\{x\in \mathbb{F}^n:Gx^\top=0\}$ (only need to check basis of $C$) 2. $C^{\perp}=\{xH|x\in \mathbb{F}^{n-k}\}$ By rank-nullity theorem, $dim(C^{\perp})=n-dim(C)=n-k$. > [!WARNING] > > $C^{\perp}\cap C=\{0\}$ is not always true. > > Let $C=\{(0,0),(1,1)\}\subseteq \mathbb{F}_2^2$. Then $C^{\perp}=\{(0,0),(1,1)\}\subseteq \mathbb{F}_2^2$ since $(1,1)\begin{pmatrix} 1\\ 1\end{pmatrix}=0$. ### Example of binary codes #### Trivial code Let $\mathbb{F}=\mathbb{F}_2$. Let $C=\mathbb{F}^n$. Generator matrix is identity matrix. Parity check matrix is zero matrix. Minimum distance is 1. #### Parity code Let $\mathbb{F}=\mathbb{F}_2$. Let $C=\{(c_1,c_2,\cdots,c_{k},\sum_{i=1}^k c_i)\}$. The generator matrix is: $$ G=\begin{pmatrix} 1 & 0 & 0 & \cdots & 0 & 1\\ 0 & 1 & 0 & \cdots & 0 & 1\\ 0 & 0 & 1 & \cdots & 0 & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & 1 \end{pmatrix} $$ The parity check matrix is: $$ H=\begin{pmatrix} 1 & 1 & 1 & \cdots & 1 & 1 \end{pmatrix} $$ Minimum distance is 2. $C^{\perp}$ is the repetition code. #### Lemma for minimum distance The minimum distance of $\mathcal{C}$ is the maximum integer $d$ such that every $d-1$ columns of $H$ are linearly independent.
Proof Assume minimum distance is $d$. Show that every $d-1$ columns of $H$ are independent. - Fact: In linear codes minimum distance is the minimum weight ($d_H(x,y)=w_H(x-y)$). Indeed, if there exists a $d-1$ columns of $H$ that are linearly dependent, then we have $Hc^\top=0$ for some $c\in \mathcal{C}$ with $w_H(c) #### The Hamming code Let $m\in \mathbb{N}$. Take all $2^m-1$ non-zero vectors in $\mathbb{F}_2^m$. Put them as columns of a matrix $H$. Example: for $m=3$, $$ H=\begin{pmatrix} 1 & 0 & 0 & 1 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 & 0 & 1 & 1\\ 0 & 0 & 1 & 0 & 1 & 1 & 1\\ \end{pmatrix} $$ Minimum distance is 3.
Proof for minimum distance Using the lemma for minimum distance. Since $H$ are linearly independent, the minimum distance is 3.
So the maximum number of error correction is 1. The length of code is $2^m-1$. $k=2^m-m-1$. #### Hadamard code Define the code by encoding function: $E(x): \mathbb{F}_2^m\to \mathbb{F}_2^{2^m}=(xy_1^\top,\cdots,xy_{2^m}^\top)$ ($y\in \mathbb{F}_2^m$) Space of codewords is image of $E$. This is a linear code since each term is a linear combination of $x$ and $y$. If $x_1,x_2,\ldots,x_m$ is a basis of $\mathbb{F}_2^m$, then $E(x_1),E(x_2),\ldots,E(x_m)$ is a basis of $\mathcal{C}$. So the dimension of $\mathcal{C}$ is $m$. Minimum distance is: $2^{m-1}$.
Proof for minimum distance Since the code is linear, then the minimum distance is the minimum weight of the codewords. For each $x\in \mathbb{F}_2^m$, there exists $2^{m-1}$ such that $E(x)=1$. For all non-zero $x$ we have $d(E(x),E(0))=2^{m-1}$.
There exists a redundant $y_i=0$, we remove it from $E(x)$ to get a puntured Hadamard codeword. The length of the code is $2^{m-1}$. SO $E: \mathbb{F}_2^m\to \mathbb{F}_2^{2^{m-1}}$ is a linear code. The generator matrix is the parity check matrix of Hamming code. The dual of Hamming code is the (puntured) Hadamard code.