Linear Codes : Error detection

[mathmari]

Hey!

The code words of a linear code $C$ have the length $n=5$.

Writing the code words into a matrix to get the linear independent ones, we get the following:
\begin{equation*}\begin{pmatrix}0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0\end{pmatrix}\rightarrow \ldots \rightarrow \begin{pmatrix}1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{pmatrix}\end{equation*}

So the dimension of $C$ is $m=2$.

Wir have also the minimum distance $d(C) =3$.

The generator matrix of $C$ is \begin{equation*}G=\begin{pmatrix}1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 \end{pmatrix}\end{equation*}

The canonical generator matrix is \begin{equation*}G'=\begin{pmatrix}1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 \end{pmatrix}\end{equation*}

And the canonical parity check matrix is
\begin{equation*}H'=\begin{pmatrix}1 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1\end{pmatrix}\end{equation*} Now there is the following question:

How many errors are at least detected at the code $C$ if $11100$ is received?

Could you give me a hint for that? (Wondering)

 

[mmwave]

Hi there! It seems like you have a good understanding of linear codes and their properties. To answer your question, we first need to understand the concept of error detection in a linear code.

In a linear code, errors can occur during transmission of a code word. These errors can change the code word to a different one, which is called a codeword error. The minimum distance $d(C)$ of a code is the minimum number of errors that can occur during transmission without changing the code word to another valid one. This means that any error pattern with $d(C)-1$ or fewer errors can be detected by the code, while any pattern with $d(C)$ or more errors will result in a valid code word.

In your example, the received code word is $11100$. This can be interpreted as a linear combination of the rows of the canonical parity check matrix $H'$:
\begin{equation*} 11100 = 1 \cdot \begin{pmatrix}1 \\ 1 \\ 1 \\ 0 \\ 0\end{pmatrix} + 1 \cdot \begin{pmatrix}1 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix} + 1 \cdot \begin{pmatrix}1 \\ 1 \\ 0 \\ 0 \\ 1\end{pmatrix} \end{equation*}
This means that there are three errors in the received code word, since it is a linear combination of three rows of the canonical parity check matrix. Therefore, the minimum number of errors that can be detected by the code $C$ is $d(C)-1=2$.

I hope this helps! Keep up the good work with linear codes. :)