\section{Error Detection}

An ISBN\footnote{International Standard Book Number} is a unique numeric book identifier. It comes in two forms: ISBN-10 and ISBN-13. Naturally, ISBN-10s have ten digits, and ISBN-13s have thirteen. The final digit in both versions is a \textit{check digit}.

\vspace{3mm}

Say we have a sequence of nine digits, forming a partial ISBN-10: $n_1 n_2 ... n_9$. \\
The final digit, $n_{10}$, is calculated as follows:

$$
	\Biggr( \sum_{i = 1}^{9} i \times n_i \Biggl) \text{ mod } 11
$$

If $n_{10}$ is equal to 10, it is written as \texttt{X}.


\problem{}
Which of the following could be valid ISBNs?

\begin{itemize}
	\item \texttt{0-134-54896-2}
	\item \texttt{0-307-29206-3}
	\item \texttt{0-316-00395-6}
\end{itemize}

\begin{solution}
	Only the first has an inconsistent check digit.
\end{solution}

\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}

	$10n_1 + 9n_2 + ... + 2n_9 + n_{10} \equiv$ \\
	$(-n_1) + (-2n_2) + ... + (-9n_9) + n_{10} =$ \\
	$-n_{10} + n_{10} \equiv 0$

	\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? \\
Provide an example or a proof.

\begin{solution}
	Let $S$ be the sum $10n_1 + 9n_2 + ... + 2n_9 + n_{10}$, before any digits are changed.

	\vspace{3mm}

	If you change one digit of the ISBN, $S$ changes by $km$, where $k \in \{1,2,...,10\}$ and $|m| \leq 10$. \\
	$k$ and $m$ cannot be divisible by 11, thus $km$ cannot be divisible by 11.

	\vspace{3mm}

	We know that $S \equiv 0 \text{ (mod 11)}$. \\
	After the change, the checksum is $S + km \equiv km \not\equiv 0 \text{ (mod 11)}$.
\end{solution}

\vfill

\problem{}
Take a valid ISBN-10 and swap two adjacent digits. When will the result be a valid ISBN-10? \\
This is called a \textit{transposition error}.


\begin{solution}
	Let $n_1n_2...n_{10}$ be a valid ISBN-10. \\
	When we swap $n_i$ and $n_{i+1}$, we subtract $n_i$ and add $n_{i+1}$ to the checksum.

	\vspace{3mm}

	If the new ISBN is to be valid, we must have that $n_{i+1} - n_i \equiv 0 \text{ (mod 11)}$. \\
	This is impossible unless $n_i = n_{i+1}$. Figure out why yourself.
\end{solution}

\vfill
\pagebreak

\problem{}
ISBN-13 error checking is slightly different. Given a partial ISBN-13 $n_1 n_2 n_3 ... n_{12}$, the final digit is given by

$$
 n_{13} = \Biggr[ \sum_{i=1}^{12} n_i \times (2 + (-1)^i) \Biggl] \text{ mod } 10
$$

What is the last digit of the following ISBN-13? \\
\texttt{978-0-380-97726-?}

\begin{solution}
	The final digit is 0.
\end{solution}

\vfill

\problem{}
Take a valid ISBN-13 and change one digit. Is it possible that you get another valid ISBN-13? \\
Provide an example or a proof.

\begin{solution}
	Let $n_1n_2...n_{13}$ be a valid ISBN-13. Choose some $n_i$ and change it to $m_i$. \\

	\vspace{3mm}

	Since $n_i$, $m_i$ $\in \{0, 1, 2, ..., 9\}$, $-9 \leq n_i - m_i \leq 9$. \\

	\vspace{2mm}

	Case 0: $i$ is 13 \\
	This is trivial.

	\vspace{2mm}

	Case 1: $i$ is odd \\
	For the new ISBN to be valid, we need $n_i - m_i \equiv 0 \text{ (mod 10)}$. \\
	This cannot happen if $n_i \neq m_i$.

	\vspace{2mm}

	Case 2: $i$ is even \\
	For the new ISBN to be valid, we need $3(n_i - m_i) \equiv 0 \text{ (mod 10)}$ \\
	This cannot happen, 10 and 3 are coprime.
\end{solution}

\vfill

\problem{}
Take a valid ISBN-13 and swap two adjacent digits. When will the result be a valid ISBN-13? \\
\hint{The answer here is more interesting than it was last time.}

\begin{solution}
	Say we swap $n_i$ and $n_{i+1}$, where $i \in \{1, 2, ..., 11\}$. \\
	The checksum changes by $2(n_{i+1} - n_i)$, and will \\
	remain the same if this value is $\equiv 0 \text{ (mod 10)}$.
\end{solution}

\vfill

\problem{}
\texttt{978-0-08-2066-46-6} was a valid ISBN until I changed a single digit. \\
Can you tell me which digit I changed?

\begin{solution}
	Nope, unless you look at the meaning of each digit in the spec. \\
	If you're unlucky, maybe not even then.
\end{solution}


\vfill
\pagebreak