Mark/handouts
Mark/handouts
1
0
Fork 0
Code Pull Requests 1 Actions Packages Releases Activity

ECC edits

Browse Source
This commit is contained in:
Mark 2023-06-20 10:07:35 -07:00
parent 20c0a84a0c
commit da57a6a8ea
Signed by: Mark
GPG Key ID: AD62BB059C2AAEE4
4 changed files with 131 additions and 111 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

6
Advanced/Error-Correcting Codes/main.tex
View File

@ -5,16 +5,14 @@
singlenumbering
]{../../resources/ormc_handout}
\usepackage{hyperref}
\usepackage{tikz}
\usetikzlibrary{patterns}
\uptitlel{Advanced 2}
\uptitler{Fall 2022}
\title{Error-Correcting Codes}
\subtitle{
Based on a handout by Yingkun Li \\
Revised by Mark on \today
}
\subtitle{Revised by Mark on \today}
\begin{document}

48
Advanced/Error-Correcting Codes/parts/00 detection.tex
View File

@ -30,29 +30,6 @@ Which of the following could be valid ISBNs?
\vfill
\pagebreak
\problem{}
Show that the following sum is divisible by 11 iff $n_1n_2...n_{10}$ is a valid ISBN-10.
$$
\sum_{i = 1}^{10} (11 - i)n_i
$$
\begin{solution}
Proof that valid $\implies$ divisible, working in mod 11:
\vspace{2mm}
$10n_1 + 9n_2 + ... + 2n_9 + n_{10} \equiv$ \par
$(-n_1) + (-2n_2) + ... + (-9n_9) + n_{10} =$ \par
$-n_{10} + n_{10} \equiv 0$
\vspace{2mm}
Having done this, the rest is easy. Work in reverse, or note that each step above is an iff.
\end{solution}
\vfill
\problem{}
Take a valid ISBN-10 and change one digit. Is it possible that you get another valid ISBN-10? \par
Provide an example or a proof.
@ -89,6 +66,31 @@ This is called a \textit{transposition error}.
\end{solution}
\vfill
\problem{}
Show that the following sum is divisible by 11 iff $n_1n_2...n_{10}$ is a valid ISBN-10.
$$
\sum_{i = 1}^{10} (11 - i)n_i
$$
\begin{solution}
Proof that valid $\implies$ divisible, working in mod 11:
\vspace{2mm}
$10n_1 + 9n_2 + ... + 2n_9 + n_{10} \equiv$ \par
$(-n_1) + (-2n_2) + ... + (-9n_9) + n_{10} =$ \par
$-n_{10} + n_{10} \equiv 0$
\vspace{2mm}
Having done this, the rest is easy. Work in reverse, or note that each step above is an iff.
\end{solution}
\vfill
\pagebreak
\problem{}

171
Advanced/Error-Correcting Codes/parts/01 correction.tex
View File

@ -8,7 +8,7 @@ QR codes feature a system that does. \par
Odds are, you've seen a QR code with an image in the center. Such codes aren't \say{special}---they're simply missing their central pixels. The error-correcting algorithm in the QR specification allows us to read the code despite this damage.
\begin{figure}[h]
\centering
\includegraphics[width = 3cm]{qr}
\href{https://youtube.com/watch?v=dQw4w9WgXcQ}{\includegraphics[width = 3cm]{qr}}
\end{figure}
\definition{Repeating codes}
@ -24,7 +24,7 @@ $$
If we flip any one bit, we can easily find and fix the error.
\problem{}
\problem{}<number-repeat>
How many repeated digits do you need to...
\begin{itemize}
\item[-] detect a transposition error?
@ -42,90 +42,93 @@ $$
For example, the efficiency of the three-repeat code above is $\frac{3}{9} = \frac{1}{3} \approx 0.33$
\problem{}
\problem{}<k-efficiency>
What is the efficiency of a $k$-repeat code?
\vfill
As you just saw, repeat codes are not a good solution. You need many extra bits for even a small amount of redundancy. We need a better system.
\pagebreak
\definition{Hamming's Square Code}
We will now analyze a more efficient coding scheme: \par
\vspace{1mm}
Take a four-bit message and arrange it in a $2 \times 2$ square. \par
Compute the pairity of each row and write it at the right. \par
Compute the pairity of each column and write it at the bottom. \par
Finally, compute the pairity of the entire message write it in the lower right corner.
This ensures that the total number of ones in the message is even.
\vspace{2mm}
Reading the result row by row to get the encoded message. \par
For example, the message 1011 generates the sequence 101110011:
$$
1011
\longrightarrow
\begin{array}{cc|}
1 & 0 \\
1 & 1 \\
\hline
\end{array}
\longrightarrow
\begin{array}{cc|c}
1 & 0 & 1 \\
1 & 1 & 0 \\ \hline
0 & 1 &
\end{array}
\longrightarrow
\begin{array}{cc|c}
1 & 0 & 1 \\
1 & 1 & 0 \\ \hline
0 & 1 & 1
\end{array}
\longrightarrow
101110011
$$
\problem{}
The following messages are encoded using the method above.
Find and correct any single-digit or transposition errors.
\begin{enumerate}
\item \texttt{110 110 011} %101110011
\item \texttt{100 101 011} %110101011
\item \texttt{001 010 110} %000110110
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item \texttt{101 110 011} or \texttt{110 101 011}
\item \texttt{110 101 011}
\item \texttt{000 110 110}
\end{enumerate}
\end{solution}
\vfill
\problem{}
What is the efficiency of this coding scheme?
\vfill
\problem{}
Can we correct a single-digit error in the encoded message? \par
Can we correct a transposition error in the encoded message?
\vfill
\problem{}
Let's generalize this coding scheme to a non-square table: \par
Given a message of length $ab$, construct a rectangle with dimensions $a \times b$ as described above.
\begin{itemize}
\item What is the efficiency of a $a \times b$ rectangle code?
\item Can the $a \times b$ rectangle code detect and fix single-bit errors?
\item Can the $a \times b$ rectangle code detect and fix two-bit errors?
\end{itemize}
\vfill
\pagebreak
%\definition{Hamming's Square Code}
%We will now analyze a more efficient coding scheme: \par
%
%\vspace{1mm}
%
%Take a four-bit message and arrange it in a $2 \times 2$ square. \par
%Compute the pairity of each row and write it at the right. \par
%Compute the pairity of each column and write it at the bottom. \par
%Finally, compute the pairity of the entire message write it in the lower right corner.
%This ensures that the total number of ones in the message is even.
%
%\vspace{2mm}
%
%Reading the result row by row to get the encoded message. \par
%For example, the message 1011 generates the sequence 101110011:
%
%$$
%1011
%\longrightarrow
%\begin{array}{cc|}
% 1 & 0 \\
% 1 & 1 \\
% \hline
%\end{array}
%\longrightarrow
%\begin{array}{cc|c}
% 1 & 0 & 1 \\
% 1 & 1 & 0 \\ \hline
% 0 & 1 &
%\end{array}
%\longrightarrow
%\begin{array}{cc|c}
% 1 & 0 & 1 \\
% 1 & 1 & 0 \\ \hline
% 0 & 1 & 1
%\end{array}
%\longrightarrow
%101110011
%$$
%
%\problem{}
%The following messages are encoded using the method above.
%Find and correct any single-digit or transposition errors.
%\begin{enumerate}
% \item \texttt{110 110 011} %101110011
% \item \texttt{100 101 011} %110101011
% \item \texttt{001 010 110} %000110110
%\end{enumerate}
%
%\begin{solution}
% \begin{enumerate}
% \item \texttt{101 110 011} or \texttt{110 101 011}
% \item \texttt{110 101 011}
% \item \texttt{000 110 110}
% \end{enumerate}
%\end{solution}
%
%\vfill
%
%\problem{}
%What is the efficiency of this coding scheme?
%
%\vfill
%
%\problem{}
%Can we correct a single-digit error in the encoded message? \par
%Can we correct a transposition error in the encoded message?
%
%\vfill
%
%\problem{}
%Let's generalize this coding scheme to a non-square table: \par
%Given a message of length $ab$, construct a rectangle with dimensions $a \times b$ as described above.
%\begin{itemize}
% \item What is the efficiency of a $a \times b$ rectangle code?
% \item Can the $a \times b$ rectangle code detect and fix single-bit errors?
% \item Can the $a \times b$ rectangle code detect and fix two-bit errors?
%\end{itemize}
%
%\vfill
%\pagebreak

17
Advanced/Error-Correcting Codes/parts/02 hamming.tex
View File

@ -519,6 +519,23 @@ Does this correlate with that message bit's index?
Say we have a message with exactly one single-bit error. \par
If we know which parity bits are inconsistent, how can we find where the error is?
\vfill
\pagebreak
\problem{}
How efficient is the 16-bit hamming code?
\vfill
\problem{}<generalize-hamming>
Can you generalize this system for messages of 4, 64, or 256 bits?
\vfill
\problem{}
How efficient is each code in \ref{generalize-hamming}? \par
What do we sacrifice for this efficiency gain?
\vfill
\pagebreak