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2023-06-20 10:07:35 -07:00
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Advanced/Error-Correcting Codes/parts/00 detection.tex
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@ -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{}