ECC edits

src/Advanced/Error-Correcting Codes/main.tex

@@ -16,6 +16,11 @@
 \title{Error-Correcting Codes}
 \subtitle{Prepared by Mark on \today}
 
+
+% TODO:
+% ISBN section is tedious, could use some work.
+% Check problem 5
+
 \begin{document}
 
 	\maketitle

src/Advanced/Error-Correcting Codes/parts/00 detection.tex

@@ -1,6 +1,8 @@
 \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}.
+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}
 
@@ -15,7 +17,8 @@ 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?
+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}
@@ -23,15 +26,16 @@ Only one of the following ISBNs is valid. Which one is it?
 \end{itemize}
 
 \begin{solution}
-	The first has an inconsistent check digit.
+	The second is valid.
 \end{solution}
 
 \vfill
 \pagebreak
 
 \problem{}
-Take a valid ISBN-10 and change one digit. Is it possible that you get another valid ISBN-10? \par
+Take a valid ISBN-10 and change one digit. \par
-Provide a proof.
+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.
@@ -50,9 +54,8 @@ Provide a proof.
 \vfill
 
 \problem{}
-Take a valid ISBN-10 and swap two adjacent digits. When will the result be a valid ISBN-10? \par
+Take a valid ISBN-10 and swap two adjacent digits. This is called a \textit{transposition error}. \par
-This is called a \textit{transposition error}.
+When will the result be a valid ISBN-10?
-
 
 \begin{solution}
 	Let $n_1n_2...n_{10}$ be a valid ISBN-10. \par
@@ -68,7 +71,8 @@ This is called a \textit{transposition error}.
 \pagebreak
 
 \definition{}
-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
+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
@@ -127,7 +131,7 @@ Take a valid ISBN-13 and swap two adjacent digits. When will the result be a val
 
 \vfill
 
-\problem{}<isbn-nocorrect>
+\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?
 

src/Advanced/Error-Correcting Codes/parts/01 correction.tex

@@ -1,6 +1,6 @@
 \section{Error Correction}
 
-As we saw in \ref{isbn-nocorrect}, the ISBN check-digit scheme does not allow us to correct errors. \par
+As we saw in \ref{isbnnocorrect}, the ISBN check-digit scheme does not allow us to correct errors. \par
 QR codes feature a system that does. \par
 
 \vspace{1mm}
@@ -8,7 +8,8 @@ QR codes feature a system that does. \par
 Odds are, you've seen a QR code with an image in the center. Such codes aren't \say{special}---they're simply missing their central pixels. The error-correcting algorithm in the QR specification allows us to read the code despite this damage.
 \begin{figure}[h]
 	\centering
-	\href{https://youtube.com/watch?v=dQw4w9WgXcQ}{\includegraphics[width = 3cm]{qr}}
+	%\href{https://youtube.com/watch?v=dQw4w9WgXcQ}{\includegraphics[width = 3cm]{qr-rick}}
+	\href{https://betalupi.com}{\includegraphics[width = 3cm]{qr-betalupi}}
 \end{figure}
 
 \definition{Repeating codes}
@@ -47,7 +48,8 @@ What is the efficiency of a $k$-repeat code?
 
 \vfill
 
-As you just saw, repeat codes are not a good solution. You need many extra bits for even a small amount of redundancy. We need a better system.
+Repeat codes are not efficient. We need to inflate our message by
+\textit{three times} if we want to correct even a single error. We need a better system.
 
 \pagebreak
 

src/Advanced/Error-Correcting Codes/parts/02 hamming.tex

@@ -1,6 +1,6 @@
 \section{Hamming Codes}
 
-Say we have a 16-bit message, for example \texttt{1011 0101 1101 1001}. \par
+Say we have an $n$-bit message, for example \texttt{1011 0101 1101 1001}. \par
 We will number its bits in binary, from left to right:
 
 
@@ -62,13 +62,14 @@ We will number its bits in binary, from left to right:
 
 
 \problem{}
-In this 16-bit message, how many message bits have an index with a one as the last digit? \par
+If we number the bits of a 16-bit message as described above, \par
+how many message bits have an index with a one as the last digit? \par
 (i.e, an index that looks like \texttt{***1})
 
 \vspace{2cm}
 
 \problem{}
-Say we number the bits in a 32-bit message as above. \par
+Now consider a 32-bit message. \par
 How many message bits have an index with a one as the $n^\text{th}$ digit? \par
 
 \vspace{2cm}
@@ -76,7 +77,10 @@ How many message bits have an index with a one as the $n^\text{th}$ digit? \par
 
 
 
-We now want a way to detect errors in our 16-bit message. To do this, we'll replace a few data bits with parity bits. This will reduce the amount of information we can send, but will also improve our error-detection capabilities.
+Now, let's come up with a way to detect errors in our 16-bit message.
+To do this, we'll replace a few data bits with parity bits.
+This will reduce the amount of information we can send,
+but will allow the receiver to detect errors in the received message.
 
 \vspace{1mm}
 
@@ -174,7 +178,8 @@ Can this coding scheme correct a single-bit error?
 \vfill
 \pagebreak
 
-We'll now add four more parity bits, in positions \texttt{0001}, \texttt{0010}, \texttt{0100}, and \texttt{1000}:
+We'll now add four more parity bits, in positions \texttt{0001}, \texttt{0010}, \texttt{0100}, and \texttt{1000}: \par
+This error-detection scheme is called the \textit{Hamming code}.
 
 \begin{center}
 \begin{tikzpicture}[scale = 1.25]
@@ -246,7 +251,7 @@ Bits \texttt{0100} and \texttt{1000} work in the same way. \par
 When counting bits in binary numbers, go from right to left.}
 
 \problem{}
-Which message bits does each parity bit cover? \par
+Which message bits does each parity bit \say{cover}? \par
 In other words, which message bits affect the value of each parity bit? \par
 
 \vspace{1mm}
@@ -358,7 +363,7 @@ Analyze this coding scheme.
 \vfill
 
 \problem{}
-Each of the following messages has either 0, 1, or two errors. \par
+Each of the following messages has either one, two, or no errors. \par
 Find the errors and correct them if possible. \par
 \hint{Bit \texttt{0000} should tell you how many errors you have.}
 
@@ -590,7 +595,7 @@ A \textit{message stream} is an infinite string of binary digits.
 \problem{}
 Show that Hamming codes do not reliably detect bit deletions: \par
 \hint{
-	Create a 17-bit message whose first 16 bits are a valid Hamming block, \par
+	Create a 17-bit message whose first 16 bits are a valid Hamming-coded message, \par
 	and which is still valid when a bit (chosen by you; not the $17^\text{th}$) is deleted.
 }
 
@@ -598,10 +603,11 @@ Show that Hamming codes do not reliably detect bit deletions: \par
 
 \problem{}
 Convince yourself that Hamming codes cannot correct insertions. \par
-Then, create a 16-bit message that...
+Then, create a 16-bit message that:
 \begin{itemize}
-	\item is a valid Hamming block, and
+	\item is itself a valid Hamming code, and
-	\item incorrectly "corrects" a single bit error when it encounters an insertion error.
+	\item incorrectly "corrects" a single bit error when it encounters an insertion error. \par
+	You may choose where the insertion occurs.
 \end{itemize}
 
 \vfill