Add "Cosa Nostra"

src/Advanced/Generating Functions/main.tex

@@ -1,7 +1,7 @@
 % use [nosolutions] flag to hide solutions.
 % use [solutions] flag to show solutions.
 \documentclass[
-	%solutions,
+	solutions,
 	singlenumbering
 ]{../../../lib/tex/handout}
 \usepackage{../../../lib/tex/macros}
@@ -19,5 +19,4 @@
 	\input{parts/01 fibonacci.tex}
 	\input{parts/02 dice.tex}
 	\input{parts/03 coins.tex}
-	\input{parts/04 bonus.tex}
 \end{document}

src/Advanced/Generating Functions/parts/01 fibonacci.tex

@@ -77,7 +77,7 @@ A \textit{rational function} $f$ is a function that can be written as a quotient
 That is, $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials.
 
 \problem{}
-Solve the equation from \ref{fibo} for $F(x)$, expressing it as a rational function.
+Solve the equation from \ref<fibo> for $F(x)$, expressing it as a rational function.
 
 \begin{solution}
 	\begin{align*}
@@ -99,8 +99,8 @@ Solve the equation from \ref{fibo} for $F(x)$, expressing it as a rational funct
 
 
 \definition{}
-\textit{Partial fraction decomposition} is an algebraic technique that works as follows: \par
+\textit{Partial fraction decomposition} is an algebreic technique that works as follows: \par
-If $p(x)$ is a polynomial of degree 1 and $a$ and $b$ are constants,
+If $p(x)$ is a polynomial and $a$ and $b$ are constants,
 we can rewrite the rational function $\frac{p(x)}{(x-a)(x-b)}$ as follows:
 \begin{equation*}
 	\frac{p(x)}{(x-a)(x-b)} = \frac{c}{x-a} + \frac{d}{x-b}
@@ -131,7 +131,7 @@ find a closed-form expression for its coefficients using partial fraction decomp
 
 \problem{}
 Using problems from the introduction and \ref{pfd}, find an expression
-for the coefficients of $F(x)$ (and thus, for the Fibonacci numbers).
+for the coefficients of $F(x)$ (and this, for the Fibonacci numbers).
 
 
 \begin{solution}

src/Advanced/Generating Functions/parts/02 dice.tex

@@ -76,7 +76,7 @@ the probability that the sum of the two dice is $k$.
 
 \problem{}
 Using generating functions, find two six-sided dice whose sum has the same
-distribution as the sum of two standard six-sided dice. \par
+distribution as the sum of two standard six-sided dice? \par
 
 That is, for any integer $k$, the number if ways that the sum of the two
 nonstandard dice rolls as $k$ is equal to the number of ways the sum of

src/Advanced/Generating Functions/parts/03 coins.tex

@@ -9,7 +9,7 @@ using pennies, nickels, dimes, quarters and half-dollars?}
 \vspace{2mm}
 
 Most ways of solving this involve awkward brute-force
-approaches that don't reveal anything interesting about the problem:
+approache that don't reveal anything interesting about the problem:
 how can we change our answer if we want to make change for
 \$0.51, or \$1.05, or some other quantity?
 

src/Advanced/Generating Functions/parts/04 bonus.tex

@@ -1,57 +0,0 @@
-\section{Extra Problems}
-
-
-\problem{USAMO 1996 Problem 6}
-Determine (with proof) whether there is a subset $X$ of
-the nonnegative integers with the following property: for any nonnegative integer $n$ there is exactly
-one solution of $a + 2b = n$ with $a, b \in X$.
-(The original USAMO question asked about all integers, not just nonnegative - this is harder,
-but still approachable with generating functions.)
-
-
-\vfill
-
-\problem{IMO Shortlist 1998}
-Let $a_0, a_1, ...$ be an increasing sequence of nonnegative integers
-such that every nonnegative integer can be
-expressed uniquely in the form $a_i + 2a_j + 4a_k$,
-where $i, j, k$ are not necessarily distinct.
-
-Determine $a_1998$.
-
-
-\vfill
-
-\problem{USAMO 1986 Problem 5}
-By a partition $\pi$ of an integer $n \geq 1$, we mean here a
-representation of $n$ as a sum of one or more positive integers where the summands must be put in
-nondecreasing order. (e.g., if $n = 4$, then the partitions $\pi$ are
-$1 + 1 + 1 + 1$, $1 + 1 + 2$, $1 + 3, 2 + 2$, and $4$).
-
-
-For any partition $\pi$, define $A(\pi)$ to be the number of ones which appear in $\pi$, and define $B(\pi)$
-to be the number of distinct integers which appear in $\pi$ (e.g, if $n = 13$ and $\pi$ is the partition
-$1 + 1 + 2 + 2 + 2 + 5$, then $A(\pi) = 2$ and $B(\pi) = 3$).
-
-Show that for any fixed $n$, the sum of $A(\pi)$ over all partitions of $\pi$ of $n$ is equal to the sum of
-$B(\pi)$ over all partitions of $\pi$ of $n$.
-
-\vfill
-
-\problem{USAMO 2017 Problem 2}
-Let $m_1, m_2, ..., m_n$ be a collection of $n$ distinct positive
-integers. For any sequence of integers $A = (a_1, ..., a_n)$ and any permutation $w = w_1, ..., w_n$ of
-$m_1, ..., m_n$, define an $A$-inversion of $w$ to be a pair of entries $w_i, w_j$ with $i < j$ for which one of the
-following conditions holds:
-\begin{itemize}
-	\item $ai \geq wi > wj$
-	\item $wj > ai \geq wi$
-	\item $wi > wj > ai$
-\end{itemize}
-
-Show that for any two sequences of integers $A = (a_1, ..., a_n)$ and $B = (b_1, ..., b_n)$ and for any
-positive integer $k$, the number of permutations of $m_1, ..., m_n$ having exactly $k$ $A$-inversions is equal
-to the number of permutations of $m_1, ..., m_n$ having exactly $k$ $B$-inversions.
-(The original USAMO problem allowed the numbers $m_1, ..., m_n$ to not necessarily be distinct.)
-
-\vfill

src/Advanced/Introduction to Quantum/src/parts/01 bits.tex

@@ -230,7 +230,7 @@ The \textit{tensor product} of two vectors is defined as follows:
 That is, we take our first vector, multiply the second
 vector by each of its components, and stack the result.
 You could think of this as a generalization of scalar
-multiplication, where scalar multiplication is a
+mulitiplication, where scalar mulitiplication is a
 tensor product with a vector in $\mathbb{R}^1$:
 \begin{equation*}
 	a

src/Advanced/Mock a Mockingbird/main.tex

@@ -81,6 +81,5 @@
 	\input{parts/00 intro}
 	\input{parts/01 tmam}
 	\input{parts/02 kestrel}
-	\input{parts/03 bonus}
 
 \end{document}

src/Advanced/Mock a Mockingbird/parts/01 tmam.tex

@@ -23,7 +23,7 @@ Complete his proof.
 		\lineno{} let A           \cmnt{Let A be any any bird.}
 		\lineno{} let Cx = A(Mx)  \cmnt{Define C as the composition of A and M}
 		\lineno{} CC = A(MC)
-		\lineno{}    = A(CC)
+		\lineno{}    = A(CC)  \qed{}
 	\end{alltt}
 \end{solution}
 
@@ -46,7 +46,7 @@ Show that the laws of the forest guarantee that at least one bird is egocentric.
 		\lineno{}
 		\lineno{} ME = E     \cmnt{By definition of fondness}
 		\lineno{} ME = EE    \cmnt{By definition of M}
-		\lineno{} \thus{} EE = E
+		\lineno{} \thus{} EE = E  \qed{}
 	\end{alltt}
 \end{solution}
 
@@ -99,7 +99,7 @@ Show that if $C$ is agreeable, $A$ is agreeable.
 		\lineno{} let y so that Cy = Ey  \cmnt{Such a y must exist because C is agreeable}
 		\lineno{}
 		\lineno{} A(By) = Ey
-		\lineno{}       = D(By)
+		\lineno{}       = D(By)  \qed{}
 	\end{alltt}
 \end{solution}
 
@@ -129,20 +129,6 @@ Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$
 We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\
 Note that $x$ and $y$ may be the same bird. \\
 
-
-\problem{}
-Show that any bird that is fond of at least one bird is compatible with itself.
-
-\begin{solution}
-	\begin{alltt}
-		\lineno{} let A
-		\lineno{} let x so that Ax = x  \cmnt{A is fond of at least one other bird}
-		\lineno{} Ax = x
-	\end{alltt}
-\end{solution}
-
-\vfill
-
 \problem{}
 Show that any two birds in this forest are compatible. \\
 \begin{alltt}
@@ -158,6 +144,7 @@ Show that any two birds in this forest are compatible. \\
 \begin{solution}
 	\begin{alltt}
 		\lineno{} let A, B
+		\lineno{}
 		\lineno{} let Cx = A(Bx)  \cmnt{Composition}
 		\lineno{} let y = Cy      \cmnt{Let C be fond of y}
 		\lineno{}
@@ -165,9 +152,24 @@ Show that any two birds in this forest are compatible. \\
 		\lineno{}    = A(By)
 		\lineno{}
 		\lineno{} let x = By  \cmnt{Rename By to x}
-		\lineno{} Ax = y
+		\lineno{} Ax = y  \qed{}
 	\end{alltt}
 \end{solution}
 
+\vfill
+
+\problem{}
+Show that any bird that is fond of at least one bird is compatible with itself.
+
+\begin{solution}
+	\begin{alltt}
+		\lineno{} let A
+		\lineno{} let x so that Ax = x  \cmnt{A is fond of at least one other bird}
+		\lineno{} Ax = x  \qed{}
+	\end{alltt}
+
+	That's it.
+\end{solution}
+
 \vfill
 \pagebreak

src/Advanced/Mock a Mockingbird/parts/02 kestrel.tex

@@ -18,7 +18,7 @@ Say $A$ is fixated on $B$. Is $A$ fond of $B$?
 	\begin{alltt}
 		\lineno{} let A
 		\lineno{} let B so that Ax = B
-		\lineno{} \thus{} AB = B
+		\lineno{} \thus{} AB = B \qed{}
 	\end{alltt}
 \end{solution}
 \vfill
@@ -39,7 +39,7 @@ Show that an egocentric Kestrel is hopelessly egocentric.
 	\begin{alltt}
 		\lineno{} KK = K
 		\lineno{} \thus{} (KK)y = K  \cmnt{By definition of the Kestrel}
-		\lineno{} \thus{} Ky = K     \cmnt{By 01}
+		\lineno{} \thus{} Ky = K \qed{}  \cmnt{By 01}
 	\end{alltt}
 \end{solution}
 
@@ -64,7 +64,7 @@ Given the Law of Composition and the Law of the Mockingbird, show that at least
 	\begin{alltt}
 		\lineno{} let A so that KA = A   \cmnt{Any bird is fond of at least one bird}
 		\lineno{} (KA)y = y              \cmnt{By definition of the kestrel}
-		\lineno{} \thus{} Ay = A              \cmnt{By 01}
+		\lineno{} \thus{} Ay = A \qed{}  \cmnt{By 01}
 	\end{alltt}
 \end{solution}
 
@@ -90,7 +90,7 @@ Show that $Kx = Ky \implies x = y$.
 		\lineno{} (Kx)z = x
 		\lineno{} (Ky)z = y
 		\lineno{}
-		\lineno{} \thus{} x = (Kx)z = (Ky)z = y
+		\lineno{} \thus{} x = (Kx)z = (Ky)z = y \qed{}
 	\end{alltt}
 \end{solution}
 
@@ -128,7 +128,7 @@ An egocentric Kestrel must be extremely lonely. Why is this?
 		\lineno{} Ky = K
 		\lineno{} Kx = Ky
 		\lineno{} x = y for all x, y  \cmnt{By \ref{leftcancel}}
-		\lineno{} x = y = K  \cmnt{By 10, and since K exists}
+		\lineno{} x = y = K \qed{}  \cmnt{By 10, and since K exists}
 	\end{alltt}
 \end{solution}
 

src/Advanced/Mock a Mockingbird/parts/03 bonus.tex

@@ -1,102 +0,0 @@
-\section{Bonus Problems}
-
-\definition{}
-The identity bird has sometimes been maligned, owing to
-the fact that whatever bird x you call to $I$, all $I$ does is to echo
-$x$ back to you.
-
-\vspace{2mm}
-
-Superficially, the bird $I$ appears to have no intelligence or imagination; all it can do is repeat what it hears.
-For this reason, in the past, thoughtless students of ornithology
-referred to it as the idiot bird. However, a more profound or-
-nithologist once studied the situation in great depth and dis-
-covered that the identity bird is in fact highly intelligent! The
-real reason for its apparently unimaginative behavior is that it
-has an unusually large heart and hence is fond of every bird!
-When you call $x$ to $I$, the reason it responds by calling back $x$
-is not that it can't think of anything else; it's just that it wants
-you to know that it is fond of $x$!
-
-\vspace{2mm}
-
-Since an identity bird is fond of every bird, then it is also
-fond of itself, so every identity bird is egocentric. However,
-its egocentricity doesn't mean that it is any more fond of itself
-than of any other bird!.
-
-
-\problem{}
-The laws of the forest no longer apply.
-
-Suppose we are told that the forest contains an identity bird
-$I$ and that $I$ is agreeable. \
-Does it follow that every bird must be fond of at least one bird?
-
-\vfill
-
-
-\problem{}
-Suppose we are told that there is an identity bird $I$ and that
-every bird is fond of at least one bird. \
-Does it necessarily follow that $I$ is agreeable?
-
-\vfill
-\pagebreak
-
-
-\problem{}
-Suppose we are told that there is an identity bird $I$, but we are
-not told whether $I$ is agreeable or not.
-
-However, we are told that every pair of birds is compatible. \
-Which of the following conclusiens can be validly drawn?
-
-\begin{itemize}
-	\item Every bird is fond of at least one bird
-	\item $I$ is agreeable.
-\end{itemize}
-
-\vfill
-
-\problem{}
-The identity bird $I$, though egocentric, is in general not hope-
-lessly egocentric. Indeed, if there were a hopelessly egocentric
-identity bird, the situation would be quite sad. Why?
-
-\vfill
-
-\definition{}
-A bird $L$ is called a lark if the following
-holds for any birds $x$ and $y$:
-
-\[
-	(Lx)y = x(yy)
-\]
-
-\problem{}
-Prove that if the forest contains a lark $L$ and an identity bird
-$I$, then it must also contain a mockingbird $M$.
-
-\vfill
-\pagebreak
-
-
-\problem{}
-Why is a hopelessly egocentric lark unusually attractive?
-
-\vfill
-
-
-\problem{}
-Assuming that no bird can be both a lark and a kestrel---as
-any ornithologist knows!---prove that it is impossible for a
-lark to be fond of a kestrel.
-
-\vfill
-
-\problem{}
-It might happen, however, that a kestrel is fond of a lark. \par
-Show that in this case, \textit{every} bird is fond of the lark.
-
-\vfill

src/Advanced/Retrograde Analysis/parts/03 medium.tex

@@ -251,7 +251,7 @@ What is it, and what is its color? \par
 	\textbf{Part 4:}
 
 	The promoted black bishop on H2 must have been promoted on G1. The pawn which was promoted must have come from G7,
-	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accounted for).
+	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accouted for).
 	The Pawn from E7 has promoted to the bishop on A2.
 
 	What happened was this: the white pawn from G2 made its capture on H3 while the pawn on G3 was still on H2. This allowed the black pawn

src/Intermediate/Instant Insanity/main.tex

@@ -331,7 +331,7 @@
 	representing all four cubes. \\
 
 	\begin{center} \begin{small}
-	\begin{tikzpicture} \label{pic:II_configuration}
+	\begin{tikzpicture} \label{pic:II_comfiguration}
 		\filldraw [blue] (0,5) -- (1,5) -- (1,6) --
 		(0,6) -- (0,5);
 		\draw [line width = 1.5pt] (0,5) --

src/Warm-Ups/Bugs/main.typ

@@ -1,11 +0,0 @@
-#import "@local/handout:0.1.0": *
-
-#show: handout.with(
-  title: [Warm-Up: Bugs on a Log],
-  by: "Mark",
-)
-
-#problem()
-2013 bugs are on a meter-long line. Each walks to the left or right at a constant speed. \
-If two bugs meet, both turn around and continue walking in opposite directions. \
-What is the longest time it could take for all the bugs to walk off the end of the log?

src/Warm-Ups/Bugs/meta.toml

@@ -1,6 +0,0 @@
-[metadata]
-title = "Bugs on a Log"
-
-[publish]
-handout = true
-solutions = true