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Fixed errors

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Mark
2023-06-25 18:28:08 -07:00
parent 2e18427669
commit a87e4417ce
2 changed files with 165 additions and 155 deletions
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34
Advanced/Error-Correcting Codes/parts/00 detection.tex
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@ -5,26 +5,25 @@ An ISBN\footnote{International Standard Book Number} is a unique numeric book id
\vspace{3mm}
Say we have a sequence of nine digits, forming a partial ISBN-10: $n_1 n_2 ... n_9$. \par
The final digit, $n_{10}$, is calculated as follows:
The final digit, $n_{10}$, is chosen from $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ so that:
$$
\Biggr( \sum_{i = 1}^{9} (11 - i) \times n_i \Biggl) \text{ mod } 11
\sum_{i = 1}^{10} (11 - i)n_i
$$
If $n_{10}$ is equal to 10, it is written as \texttt{X}.
\problem{}
Which of the following could be valid ISBNs?
Only one of the following ISBNs is valid. Which one is it?
\begin{itemize}
\item \texttt{0-134-54896-2}
\item \texttt{0-895-77258-2}
\item \texttt{0-316-00395-6}
\end{itemize}
\begin{solution}
Only the first has an inconsistent check digit.
The first has an inconsistent check digit.
\end{solution}
\vfill
@ -66,31 +65,6 @@ 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{}