fmt

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
-If $p(x)$ is a polynomial of degree 1 and $a$ and $b$ are constants,
+\textit{Partial fraction decomposition} is an algebreic technique that works as follows: \par
+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/Adders/main.typ

@@ -46,10 +46,10 @@ Use two half adders to construct a full adder.
 
 #solution([
   $
-    s_1, c_1 & = "HA"(a, b) \
-    s_2, c_2 & = "HA"(s_1, c_"in") \
-     s_"out" & = s_2 \
-     c_"out" & = "OR"(c_1, c_2)
+    s_1, c_1 &= "HA"(a, b) \
+    s_2, c_2 &= "HA"(s_1, c_"in") \
+    s_"out" &= s_2 \
+    c_"out" &= "OR"(c_1, c_2)
   $
 
   #v(2mm)

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

src/Warm-Ups/Cosa Nostra/main.typ

@@ -1,50 +0,0 @@
-#import "@local/handout:0.1.0": *
-
-#show: handout.with(
-  title: [Warm-Up: Cosa Nostra],
-  by: "Mark",
-)
-
-
-#problem()
-There are 36 gangsters in a certain district of Chicago.
-Some pairs of gangsters have feuds.
-
-- Each gangster is part of at least one outfit, and no two outfits share the same members.
-- If two gangsters are both in one outfit, there is no feud between them.
-- A gangster that is not in a certain outfit must have a feud with at least one if its members.
-
-What is the maximum number of outfits that can exist in this district?
-
-
-
-#solution[
-  *Definition:* Let the _authority_ of a gangster be the number of oufits they are a part of.
-
-  #v(5mm)
-
-  *Lemma:* Say two gangsters have a feud. Label them $x$ and $y$ so that $#text(`authority`) (x) > #text(`authority`) (y)$. \
-  Then, replacing $y$ with a clone of $x$ will strictly increase the number of outfits. \
-  If $#text(`authority`) (x) = #text(`authority`) (y)$, replacing $y$ with $x$ will not change the number of outfits.
-
-  #v(5mm)
-
-  *Proof:*
-  Let $A$ be the set of outfits that $y$ is a part of, and $B$ its complement (that is, all outfits that $a$ is _not_ a part of). If we delete $y$...
-  - all outfits in $B$ remain outfits.
-  - some outfits in $A$ cease to be outfits (as they are no longer maximal)
-
-  Also, no new outfits are formed. If a new outfit $o$ contains any enemies of $y$, it existed previously and is a member of $B$. If $o$ contains no enemies of $y$, it must have contained $y$ prior to deletion, and is thus a member of $A$. Therefore, the number of outfits is reduced by at most `authority(y)` when $y$ is deleted.
-
-  If we add a clone of $x$ after deleting $y$ (this clone has a feud with $x$), All previous outfits remain outfits, and `authority(x)` new outfits are created.
-  Therefore, replacing $y$ with a clone of $x$ strictly increases the number of outfits that exist.
-  We thus conclude that in the maximal case, all pairs of feuding gangsters have equal authority.
-
-  #v(5mm)
-
-  *Solution:* Consider an arbitrary gangster $g$. By the previous lemma, we can replace all gangsters $g$ has a feud with clones of itself. Repeat this for all gangsters, and we are left with groups of feuding gangsters who are friends with everyone outside their group. The total number of outfits is the product of the sizes of these groups.
-
-  #v(5mm)
-
-  This problem is now equivalent to the "Partition Products" warm-up. We want the list of numbers whose sum is 36 and whose product is maximal. The solution is to form 12 groups of three gangsters for a total of $3^12$ outfits.
-]

src/Warm-Ups/Cosa Nostra/meta.toml

@@ -1,6 +0,0 @@
-[metadata]
-title = "Cosa Nostra"
-
-[publish]
-handout = true
-solutions = true

src/Warm-Ups/Partition Products/main.typ

@@ -17,9 +17,7 @@ Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
   Of course, all $a_i$ should be greater than $1$. \
   Also, all $a_i$ should be smaller than four, since $x <= x(x-2)$ if $x >= 4$. \
   Thus, we're left with sequences that only contain 2 and 3. \
-  #note(
-    [Note that two twos are the same as one four, but we exclude fours for simplicity.],
-  )
+  #note([Note that two twos are the same as one four, but we exclude fours for simplicity.])
 
   #v(2mm)