ECC edits

.gitignore

@@ -9,6 +9,7 @@ __pycache__
 /output
 /output.zip
 *.pdf
+/manual
 
 # TeX build files
 *.synctex*

lib/typst/local/handout/0.1.0/header.typ

@@ -29,6 +29,26 @@
   )
 }
 
+
+#let short_solution_warning() = {
+  set text(ored)
+  align(
+    center,
+    block(
+      width: 60%,
+      height: auto,
+      breakable: false,
+      fill: rgb(255, 255, 255),
+      stroke: ored + 2pt,
+      inset: 3mm,
+      (
+        align(center, text(weight: "bold", size: 12pt, [Instructor's Handout]))
+          + parbreak()
+      ),
+    ),
+  )
+}
+
 #let make_header(
   title,
   subtitle: none,

lib/typst/local/handout/0.1.0/lib.typ

@@ -32,6 +32,7 @@
   title: none,
   by: none,
   subtitle: none,
+  short_warning: false,
 ) = {
   set page(
     margin: 20mm,
@@ -92,7 +93,7 @@
 
   // Make handout title
   {
-    import "header.typ": make_header, solution_warning
+    import "header.typ": make_header, solution_warning, short_solution_warning
     import "solution.typ": show_solutions
 
     let url = link(
@@ -108,8 +109,13 @@
       top_right: url,
     )
 
+
     if show_solutions {
-      solution_warning()
+      if short_warning {
+        short_solution_warning()
+      } else {
+        solution_warning()
+      }
     }
   }
 

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
-Provide a proof.
+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.
@@ -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
-This is called a \textit{transposition error}.
-
+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
@@ -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

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

@@ -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 incorrectly "corrects" a single bit error when it encounters an insertion error.
+	\item is itself a valid Hamming code, and
+	\item incorrectly "corrects" a single bit error when it encounters an insertion error. \par
+	You may choose where the insertion occurs.
 \end{itemize}
 
 \vfill

src/Advanced/Gods, Demons, Mortals/main.typ

@@ -4,6 +4,7 @@
   title: [Gods, Demons, and Mortals],
   by: "Mark",
   subtitle: [Based on Raymond Smullyan's _To Mock a Mockingbird_.],
+  short_warning: true,
 )
 
 #include "parts/00 warmup.typ"

src/Advanced/Gods, Demons, Mortals/parts/00 warmup.typ

@@ -3,16 +3,11 @@
 = Warm-Up
 
 #problem("The Flower Garden")
-In a certain flower garden, each flower was either red, yellow,
-or blue, and all three colors were represented. A statistician
-once visited the garden and made the observation that whatever three flowers you picked, at least one of them was bound
-to be red.
+In a certain garden, each flower was either red, yellow, or blue, and all three colors were represented. A statistician once visited the garden and made the observation that whatever three flowers you picked, at least one of them was bound to be red.
 
 #v(2mm)
 
-A second statistician visited the garden and made
-the observation that whatever three flowers you picked, at
-least one was bound to be yellow.
+A second statistician visited the garden and made the observation that whatever three flowers you picked, at least one was bound to be yellow.
 
 #v(2mm)
 
@@ -22,8 +17,11 @@ doesn't it?" The second student said: "Of course not!"
 Which student was right, and why?
 
 #solution[
-  The first student was right, and here is why. From the
-  first statistician's report it follows that there cannot be more than one yellow flower, because if there were two yellows, you could pick two yellows and one blue, thus having a group of three flowers that contained no red. This is contrary to the report that every group of three is bound to contain at least one red flower. Therefore there cannot be more than one yellow flower.
+  The first student was right.
+
+  #v(2mm)
+
+  From the first statistician's report it follows that there cannot be more than one yellow flower, because if there were two yellows, you could pick two yellows and one blue, thus having a group of three flowers that contained no red. This is contrary to the report that every group of three is bound to contain at least one red flower. Therefore there cannot be more than one yellow flower.
 
   #v(2mm)
 
@@ -31,24 +29,16 @@ Which student was right, and why?
 
   #v(2mm)
 
-  And it follows from the report of the second statistician that
-  there is at most one red flower, for if there were two reds,
-  you could pick two reds and one blue, thus obtaining a group
-  of three that contained no yellow. It also follows from the
-  second report that there cannot be more than one blue,
-  although we have already deduced this from the first report.
+  And it follows from the report of the second statistician that there is at most one red flower, for if there were two reds,
+  you could pick two reds and one blue, thus obtaining a group of three that contained no yellow. It also follows from the
+  second report that there cannot be more than one blue, although we have already deduced this from the first report.
 
   #v(2mm)
 
-  The upshot of all this is that there are only three flowers
-  in the entire garden---one red, one yellow, and one blue! And
-  so it is of course true that whatever three flowers you pick,
-  one of them must be blue.
+  The upshot of all this is that there are only three flowers in the entire garden---one red, one yellow, and one blue! And so it is of course true that whatever three flowers you pick, one of them must be blue.
 ]
 
 #v(1fr)
-#pagebreak()
-
 
 #problem("What Question")
 There is a question I could ask you that has a definite correct
@@ -62,50 +52,11 @@ Can you figure out what question I could have in mind?
 
 #solution[
   Suppose I ask you: "Is no your answer to this question?"
+
   If you answer yes, then you are affirming that no is your answer to the question,
   which is of course wrong. If you answer
   no, then you are denying that no is your answer, although no
   was your answer.
-
-  #v(2mm)
-
-  It is therefore impossible for you to answer
-  the question correctly even though the question does have a
-  correct answer: Either you answer no or you don't. If you do,
-  then yes is the correct answer; if you don't, then no is the
-  correct answer, but in neither case can you give the correct
-  answer.
 ]
 
 #v(1fr)
-
-#problem("Which Way Would You Bet?")
-Choose your favorite baseball team and consider the scores it will make next
-season. \
-Which do you bet will be the larger number---the sum
-of these scores or the product of these scores?
-
-#solution[
-  The chances are that the sum will be the larger number,
-  because your team will probably score at least one zero, and
-  one zero makes the entire product zero.
-]
-
-#v(1fr)
-
-#generic("Aside:")
-Speaking of probability and statistics, there is the story of
-a statistician who told a friend that he never took airplanes: "I
-have computed the probability that there will be a bomb on
-the plane," he explained, "and although this probability is low,
-it is still too high for my comfort."
-
-#v(2mm)
-
-Two weeks later, the friend
-met the statistician on a plane. "How come you changed your
-theory?" he asked. "Oh, I didn't change my theory; it's just
-that I subsequently computed the probability that there would
-simultaneously be two bombs on a plane. This probability is
-low enough for my comfort. \
-So now I simply carry my own bomb."

src/Advanced/Mock a Mockingbird/parts/00 intro.tex

@@ -39,7 +39,7 @@ $$
 	Cx = A(Bx)
 $$
 In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$. \\
-Note that $C$ is exactly the kind of bird the Law of Composition guarantees.
+Note that $C$ is exactly the kind of bird $L_1$ guarantees.
 
 
 \vfill

tools/scripts/build.py

@@ -5,6 +5,7 @@ import tomllib
 import shutil
 import json
 import os
+import sys
 
 # TODO:
 # list handouts without solutions
@@ -16,10 +17,22 @@ ROOT: Path = Path(os.getcwd())
 
 ### CONFIGURATION
 OUT_DIR: Path = ROOT / "output"
+OUT_DIR_MANUAL: Path = ROOT / "manual"
 TYPST_PATH: str = "typst"
 XETEX_PATH: str = "xelatex"
 ### END CONFIGURATION
 
+
+# If we're given an argument, build one handout.
+# This places output into `OUT_DIR_MANUAL`
+#
+# If we're given no arguments, build everything,
+# Placing output into `OUT_DIR`
+target = None
+if len(sys.argv) == 2:
+	target = Path(sys.argv[1])
+	print(f"Compiling `{target}`")
+
 # Allow path override
 _env = os.environ.get("TYPST_PATH")
 if isinstance(_env, str):
@@ -35,13 +48,6 @@ if not ROOT.is_dir():
 	log("Root is not a directory, cannot continue")
 	exit(1)
 
-log(f"Output dir is {OUT_DIR}")
-if OUT_DIR.exists():
-	log("Output dir exists, removing")
-	shutil.rmtree(OUT_DIR)
-
-OUT_DIR.mkdir(parents=True)
-
 
 IndexEntry = TypedDict(
 	"IndexEntry",
@@ -115,7 +121,9 @@ def log_error(res):
 	exit(1)
 
 
-def build_typst(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
+def build_typst(
+	source_dir: Path, out_subdir: Path, *, out_dir=OUT_DIR
+) -> IndexEntry | None:
 	if not (source_dir / "main.typ").is_file():
 		# log(f"No main.typ, skipping {source_dir}")
 		return None
@@ -136,7 +144,7 @@ def build_typst(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
 	# Build handout
 	log(f"Building typst (handout)  : {source_dir}")
 
-	out = OUT_DIR / out_subdir
+	out = out_dir / out_subdir
 	out.mkdir(parents=True, exist_ok=True)
 
 	res = subprocess.run(
@@ -190,7 +198,9 @@ def build_typst(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
 	}
 
 
-def build_xetex(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
+def build_xetex(
+	source_dir: Path, out_subdir: Path, *, out_dir=OUT_DIR
+) -> IndexEntry | None:
 	if not (source_dir / "main.tex").is_file():
 		# log(f"No main.tex, skipping {source_dir}")
 		return None
@@ -211,7 +221,7 @@ def build_xetex(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
 	# Build handout
 	log(f"Building xetex (handout)  : {source_dir}")
 
-	out = OUT_DIR / out_subdir
+	out = out_dir / out_subdir
 	out.mkdir(parents=True, exist_ok=True)
 
 	res = subprocess.run(
@@ -297,10 +307,31 @@ def build_dir(base: str, out_sub: str, index: list[IndexEntry]):
 				break
 
 
-index: list[IndexEntry] = []
-build_dir("src/Advanced", "Advanced", index)
-build_dir("src/Intermediate", "Intermediate", index)
-build_dir("src/Warm-Ups", "Warm-Ups", index)
+if target is None:
+	log(f"Output dir is {OUT_DIR}")
+	if OUT_DIR.exists():
+		log("Output dir exists, removing")
+		shutil.rmtree(OUT_DIR)
 
-with open(OUT_DIR / "index.json", "w") as f:
-	f.write(json.dumps(index))
+	OUT_DIR.mkdir(parents=True)
+
+	index: list[IndexEntry] = []
+	build_dir("src/Advanced", "Advanced", index)
+	build_dir("src/Intermediate", "Intermediate", index)
+	build_dir("src/Warm-Ups", "Warm-Ups", index)
+
+	with open(OUT_DIR / "index.json", "w") as f:
+		f.write(json.dumps(index))
+
+else:
+	log(f"Output dir is {OUT_DIR_MANUAL}")
+	# if OUT_DIR_MANUAL.exists():
+	# log("Output dir exists, removing")
+	# shutil.rmtree(OUT_DIR_MANUAL)
+	OUT_DIR_MANUAL.mkdir(parents=True, exist_ok=True)
+
+	builders = [build_typst, build_xetex]
+	for builder in builders:
+		res = builder(target, Path("."), out_dir=OUT_DIR_MANUAL)
+		if res is not None:
+			break