165 lines
4.2 KiB
TeX
165 lines
4.2 KiB
TeX
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\section{Error Detection}
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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}.
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\vspace{3mm}
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Say we have a sequence of nine digits, forming a partial ISBN-10: $n_1 n_2 ... n_9$. \\
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The final digit, $n_{10}$, is calculated as follows:
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$$
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\Biggr( \sum_{i = 1}^{9} i \times n_i \Biggl) \text{ mod } 11
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$$
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If $n_{10}$ is equal to 10, it is written as \texttt{X}.
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\problem{}
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Which of the following could be valid ISBNs?
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\begin{itemize}
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\item \texttt{0-134-54896-2}
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\item \texttt{0-307-29206-3}
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\item \texttt{0-316-00395-6}
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\end{itemize}
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\begin{solution}
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Only the first has an inconsistent check digit.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Show that the following sum is divisible by 11 iff $n_1n_2...n_{10}$ is a valid ISBN-10.
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$$
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\sum_{i = 1}^{10} (11 - i)n_i
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$$
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\begin{solution}
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Proof that valid $\implies$ divisible, working in mod 11:
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\vspace{2mm}
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$10n_1 + 9n_2 + ... + 2n_9 + n_{10} \equiv$ \\
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$(-n_1) + (-2n_2) + ... + (-9n_9) + n_{10} =$ \\
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$-n_{10} + n_{10} \equiv 0$
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\vspace{2mm}
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Having done this, the rest is easy. Work in reverse, or note that each step above is an iff.
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\end{solution}
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\vfill
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\problem{}
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Take a valid ISBN-10 and change one digit. Is it possible that you get another valid ISBN-10? \\
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Provide an example or a proof.
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\begin{solution}
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Let $S$ be the sum $10n_1 + 9n_2 + ... + 2n_9 + n_{10}$, before any digits are changed.
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\vspace{3mm}
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If you change one digit of the ISBN, $S$ changes by $km$, where $k \in \{1,2,...,10\}$ and $|m| \leq 10$. \\
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$k$ and $m$ cannot be divisible by 11, thus $km$ cannot be divisible by 11.
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\vspace{3mm}
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We know that $S \equiv 0 \text{ (mod 11)}$. \\
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After the change, the checksum is $S + km \equiv km \not\equiv 0 \text{ (mod 11)}$.
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\end{solution}
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\vfill
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\problem{}
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Take a valid ISBN-10 and swap two adjacent digits. When will the result be a valid ISBN-10? \\
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This is called a \textit{transposition error}.
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\begin{solution}
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Let $n_1n_2...n_{10}$ be a valid ISBN-10. \\
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When we swap $n_i$ and $n_{i+1}$, we subtract $n_i$ and add $n_{i+1}$ to the checksum.
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\vspace{3mm}
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If the new ISBN is to be valid, we must have that $n_{i+1} - n_i \equiv 0 \text{ (mod 11)}$. \\
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This is impossible unless $n_i = n_{i+1}$. Figure out why yourself.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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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
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$$
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n_{13} = \Biggr[ \sum_{i=1}^{12} n_i \times (2 + (-1)^i) \Biggl] \text{ mod } 10
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$$
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What is the last digit of the following ISBN-13? \\
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\texttt{978-0-380-97726-?}
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\begin{solution}
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The final digit is 0.
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\end{solution}
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\vfill
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\problem{}
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Take a valid ISBN-13 and change one digit. Is it possible that you get another valid ISBN-13? \\
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Provide an example or a proof.
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\begin{solution}
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Let $n_1n_2...n_{13}$ be a valid ISBN-13. Choose some $n_i$ and change it to $m_i$. \\
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\vspace{3mm}
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Since $n_i$, $m_i$ $\in \{0, 1, 2, ..., 9\}$, $-9 \leq n_i - m_i \leq 9$. \\
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\vspace{2mm}
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Case 0: $i$ is 13 \\
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This is trivial.
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\vspace{2mm}
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Case 1: $i$ is odd \\
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For the new ISBN to be valid, we need $n_i - m_i \equiv 0 \text{ (mod 10)}$. \\
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This cannot happen if $n_i \neq m_i$.
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\vspace{2mm}
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Case 2: $i$ is even \\
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For the new ISBN to be valid, we need $3(n_i - m_i) \equiv 0 \text{ (mod 10)}$ \\
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This cannot happen, 10 and 3 are coprime.
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\end{solution}
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\vfill
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\problem{}
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Take a valid ISBN-13 and swap two adjacent digits. When will the result be a valid ISBN-13? \\
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\hint{The answer here is more interesting than it was last time.}
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\begin{solution}
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Say we swap $n_i$ and $n_{i+1}$, where $i \in \{1, 2, ..., 11\}$. \\
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The checksum changes by $2(n_{i+1} - n_i)$, and will \\
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remain the same if this value is $\equiv 0 \text{ (mod 10)}$.
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\end{solution}
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\vfill
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\problem{}
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\texttt{978-0-08-2066-46-6} was a valid ISBN until I changed a single digit. \\
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Can you tell me which digit I changed?
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\begin{solution}
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Nope, unless you look at the meaning of each digit in the spec. \\
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If you're unlucky, maybe not even then.
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\end{solution}
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\vfill
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\pagebreak
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