\section{Error Detection}

An ISBN\footnote{International Standard Book Number} is a unique identifier publishers assign to their books. \par
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: $n_1 n_2 ... n_9$. \par
The final digit, $n_{10}$, is chosen from $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ so that:

$$
	\sum_{i = 1}^{10} (11 - i)n_i \equiv 0 \text{ mod } 11
$$

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


\problem{}
Only one of the following ISBNs is valid. Which one is it? \par
\note[Note]{Dashes are meaningless.}

\begin{itemize}
	\item \texttt{0-134-54896-2}
	\item \texttt{0-895-77258-2} % oliver twist
\end{itemize}

\begin{solution}
	The second is valid.
\end{solution}

\vfill
\pagebreak

\problem{}
Take a valid ISBN-10 and change one digit. \par
Is it possible that you get another valid ISBN-10? \par
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 $|m| \leq 10$. \par
	$k$ and $m$ cannot be divisible by 11, thus $km$ cannot be divisible by 11.

	\vspace{3mm}

	We know that $S \equiv 0 \text{ (mod 11)}$. \par
	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. This is called a \textit{transposition error}. \par
When will the result be a valid ISBN-10?

\begin{solution}
	Let $n_1n_2...n_{10}$ be a valid ISBN-10. \par
	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 $n_{i+1} - n_i \equiv 0 \text{ (mod 11)}$. \par
	This is impossible unless $n_i = n_{i+1}$. Figure out why yourself.
\end{solution}

\vfill
\pagebreak

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

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

\problem{}
What is the last digit of the following ISBN-13? \par
\texttt{978-030-7292-06*} % foundation

\begin{solution}
	The final digit is 3.
\end{solution}

\vfill

\problem{}
Take a valid ISBN-13 and change one digit. Is it possible that you get another valid ISBN-13? \par
If you can, provide an example; if you can't, provide a proof.

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

	\vspace{3mm}

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

	\vspace{2mm}

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

	\vspace{2mm}

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

	\vspace{2mm}

	Case 2: $i$ is even \par
	For the new ISBN to be valid, we need $3(n_i - m_i) \equiv 0 \text{ (mod 10)}$ \par
	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? \par
\hint{The answer here is more interesting than it was last time.}

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

\vfill

\problem{}<isbnnocorrect>
\texttt{978-008-2066-466} was a valid ISBN until I changed a single digit. \par
Can you find the digit I changed? Can you recover the original ISBN?

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


\vfill
\pagebreak
