Retrograde edits

Reviewed-on: #26

src/Advanced/De Bruijn/parts/0 intro.tex

@@ -21,7 +21,7 @@ Unlock this lock with only 5 keypresses.
 \end{solution}
 \vfill
 
-Now, consider the same lock, now set with a three-digit binary code.
+Now consider the same lock, but configured with a three-digit binary code.
 \problem{}
 How many codes are possible?
 \vfill

src/Advanced/De Bruijn/parts/1 words.tex

@@ -20,7 +20,11 @@ We say $v$ is a \textit{subword} of $w$ if $v$ is contained in $w$. \par
 For example, \texttt{11} is a subword of \texttt{011}, but \texttt{00} is not.
 
 \definition{}
-Recall \ref{lockproblem}. Let's generalize this to the \textit{$n$-subword problem}: \par
+Recall the lock problem from the previous page.
+Let's generalize this to the \textit{$n$-subword problem}:
+
+\vspace{1mm}
+
 Given an alphabet $A$ and a positive integer $n$,
 we want a word over $A$ that contains all possible length-$n$ subwords.
 The shortest word that solves a given $n$-subword problem is called the \textit{optimal solution}.
@@ -67,7 +71,7 @@ Find the following:
 
 \problem{}<sbounds>
 Let $w$ be a word over an alphabet of size $k$. \par
-Prove the following:
+Show that all of the following are true:
 \begin{itemize}
 	\item $\mathcal{S}_n(w) \leq k^n$
 	\item $\mathcal{S}_n(w) \geq \mathcal{S}_{n-1}(w) - 1$
@@ -103,7 +107,7 @@ Prove the following:
 
 \definition{}
 Let $v$ and $w$ be words over the same alphabet. \par
-The word $vw$ is the word formed by writing $v$ after $w$. \par
+The word $vw$ is the word formed by writing $w$ after $v$. \par
 For example, if $v = \texttt{1001}$ and $w = \texttt{10}$, $vw$ is $\texttt{100110}$.
 
 \problem{}
@@ -116,7 +120,6 @@ We'll call this the \textit{Fibonacci word} of order $k$.
 	\item What are $F_3$, $F_4$, and $F_5$?
 	\item Compute $\mathcal{S}_0$ through $\mathcal{S}_5$ for $F_5$.
 	\item Show that the length of $F_k$ is the $(k + 2)^\text{th}$ Fibonacci number. \par
-	\hint{Induction.}
 \end{itemize}
 
 \begin{solution}

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}
@@ -19,4 +19,5 @@
 	\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 algebreic technique that works as follows: \par
-If $p(x)$ is a polynomial and $a$ and $b$ are constants,
+\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,
 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 this, for the Fibonacci numbers).
+for the coefficients of $F(x)$ (and thus, 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
-approache that don't reveal anything interesting about the problem:
+approaches 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

@@ -0,0 +1,57 @@
+\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
-mulitiplication, where scalar mulitiplication is a
+multiplication, where scalar multiplication is a
 tensor product with a vector in $\mathbb{R}^1$:
 \begin{equation*}
 	a

src/Advanced/Mock a Mockingbird/main.tex

@@ -81,5 +81,6 @@
 	\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)  \qed{}
+		\lineno{}    = A(CC)
 	\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  \qed{}
+		\lineno{} \thus{} EE = E
 	\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)  \qed{}
+		\lineno{}       = D(By)
 	\end{alltt}
 \end{solution}
 
@@ -129,6 +129,20 @@ 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}
@@ -144,7 +158,6 @@ 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{}
@@ -152,24 +165,9 @@ 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  \qed{}
+		\lineno{} Ax = y
 	\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 \qed{}
+		\lineno{} \thus{} AB = B
 	\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 \qed{}  \cmnt{By 01}
+		\lineno{} \thus{} Ky = K     \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 \qed{}  \cmnt{By 01}
+		\lineno{} \thus{} Ay = A              \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 \qed{}
+		\lineno{} \thus{} x = (Kx)z = (Ky)z = y
 	\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 \qed{}  \cmnt{By 10, and since K exists}
+		\lineno{} x = y = K  \cmnt{By 10, and since K exists}
 	\end{alltt}
 \end{solution}
 

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

@@ -0,0 +1,102 @@
+\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/02 easy.tex

@@ -127,10 +127,6 @@ Mate the king in one move. \par
 \pagebreak
 
 
-
-
-
-
 % Sherlock, a question of survival
 \problem{An empty board}
 \difficulty{2}{5}
@@ -161,42 +157,6 @@ There is one more piece on the board, which isn't shown. What color square does
 \pagebreak
 
 
-% Sherlock, another monochromatic
-\problem{The knight's grave}
-\difficulty{3}{5}
-
-In the game below, no pieces have moved from a black square to a white square, or from a white square to a black square.
-The white king has made less than fourteen moves. \par
-Use this information to show that a pawn was promoted. \par
-
-% spell:off
-\manyboards{
-	ke8,
-	Pb2,Pd2,
-	Ke1
-}
-% spell:on
-
-\begin{solution}
-	Knights always move to a different colored square, so all four missing knights must have been captured on their home square.
-	What pieces captured them?
-
-	\vspace{2mm}
-
-	We can easily account for the white knights and the black knight on G8, but who could've captured the knight from B8?
-	The only white pieces that can move to black squares are pawns, the Bishop (which is trapped on C1), the rook (which is stuck on column A and row 1), or the king (which would need at least 14 moves to do so).
-
-	\vspace{2mm}
-
-	If this knight was captured by a pawn, that pawn would be immediately promoted. If it was captured by a piece that wasn't a pawn, that piece must be a promoted pawn.
-\end{solution}
-
-\vfill
-\pagebreak
-
-
-
-
 
 
 % Arabian Knights, intro (given with solution)
@@ -372,4 +332,122 @@ Which bishop was it, and what did it capture? \par
 \end{solution}
 
 \vfill
-\pagebreak
+\pagebreak
+
+
+% Sherlock, appendix
+\problem{Moriarty's first}
+\difficulty{3}{5}
+
+
+No captures have been made in the last four moves. \par
+It is White's move. What was the previous move?
+
+% spell:off
+\manyboards{
+	Bc8,
+	pg6,
+	Pg5,kh5,
+	Pd4,Qg4,Bh4,
+	pd3,
+	Pd2,Be2,Bg2,
+	Nc1,rd1,Ne1,Kf1,Qg1,Rh1
+}
+% spell:on
+
+\begin{solution}
+	To see what the position was four moves ago,
+	move the Black queen to E4, the knight on E1 to F3,
+	the Black bishop to E1, and the White bishop on C8 to H3.
+
+	The following sequence of moves brought the game to the present position:
+
+	\begin{itemize}
+		\item bishop to c8, check
+		\item bishop to h4, check
+		\item knight to e1, check
+		\item queen to g4.
+	\end{itemize}
+
+	This is the only way the present position could have arisen,
+	so Black's last move was with the queen from E4 to G4.
+
+
+	Try any other last move, and you will find it impossible to play back three more moves.
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+% Sherlock, appendix
+\problem{Moriarty's second}
+\difficulty{3}{5}
+
+
+Neither the White king nor queen has moved
+during the last five moves, nor has any piece
+been captured during that time.
+What was the last move?
+
+% spell:off
+\manyboards{
+	kh8,
+	Kg6,Bh6,
+	pa4,
+	Qa2
+}
+% spell:on
+
+\begin{solution}
+	Put the Black pawn on A7, the Black king on G8, remove the
+	White bishop, and put a White pawn on d5; this was the position
+	five moves ago. The following sequence of moves brought the
+	game to its present position:
+
+	\begin{itemize}
+		\item White: P-d6
+		\item Black: K-h8
+		\item White: P-d7
+		\item Black: P-a6
+		\item White: P-d8 = B
+		\item Black: P-a5
+		\item White: B-g5
+		\item Black: P-a4
+		\item White: B-h6
+	\end{itemize}
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+% Sherlock, appendix
+\problem{Moriarty's third}
+\difficulty{3}{5}
+
+
+No pawn has moved, nor has any piece been
+captured in the last five moves. \par
+The Black king has been accidentally
+knocked off the board. \par
+On what square should he stand?
+
+% spell:off
+\manyboards{
+	rh8,
+	pa7,pb7,pc7,pd7,pe7,Kf7,pg7,Ph7,
+	Pg6,
+	na2
+}
+% spell:on
+
+\begin{solution}
+	The only way to avoid a retrograde stalemate for White is by
+	placing the Black king on C8. Black's last move was with
+	the rook from D8, White's move before that was with his
+	king from G8, and Black's move before that was to castle.
+\end{solution}
+
+\vfill
+\pagebreak

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

@@ -169,16 +169,156 @@ White to move. Which side of the board did each color start on? \par
 
 
 
+% Sherlock, another monochromatic
+\problem{Monochromatic}
+\difficulty{4}{5}
+
+In the game below, no pieces have moved from a black square to a white square or from a white square to a black square.
+The white king has made fewer than fourteen moves. \par
+Use this information to show that a pawn was promoted. \par
+
+% spell:off
+\manyboards{
+	ke8,
+	Pb2,Pd2,
+	Ke1
+}
+% spell:on
+
+\begin{solution}
+	Knights always move to a different colored square, so all four missing knights must have been captured on their home square.
+	What pieces captured them?
+
+	\vspace{2mm}
+
+	We can easily account for the white knights and the black knight on G8, but who could've captured the knight from B8?
+	The only white pieces that can move to black squares are pawns, the Bishop (which is trapped on C1), the rook (which is stuck on column A and row 1), or the king (which would need at least 14 moves to do so).
+
+	\vspace{2mm}
+
+	If this knight was captured by a pawn, that pawn would be immediately promoted. If it was captured by a piece that wasn't a pawn, that piece must be a promoted pawn.
+\end{solution}
+
+\vfill
+\pagebreak
 
 
 
 
+% Sherlock, another question of location
+\problem{Superposition}
+\difficulty{4}{5}
+
+A white pawn is missing; it is either on F2 or G2. \par
+Where is it?
+
+% spell:off
+\manyboards{
+	ke8,rh8,
+	pa7,pf7,pg7,
+	pa6,pb6,
+	pb5,
+	Pa4,Pb4,Pc4,
+	pa3,
+	Pa2,Pb2,
+	Ke1
+}
+% spell:on
+
+\vfill
+\pagebreak
+
+% Sherlock, another question of location
+\problem{Possibility}
+\difficulty{4}{5}
+
+Show that black can castle to either side. \par
+We know the following:
+
+\begin{itemize}
+	\item White started the game missing one rook.
+	\item White has not moved either knight
+	\item No promotions have been made
+	\item White's last move was from E2 to E4.
+\end{itemize}
+
+% spell:off
+\manyboards{
+	ra8,ke8,rh8,
+	pa7,bb7,pc7,pd7,pf7,pg7,ph7,
+	nc6,nh6,
+	pe5,qg5,
+	bb4,Pe4,
+	Pb2,Pc2,Pd2,Pf2,Pg2,Ph2,
+	Nb1,Bc1,Qd1,Ke1,Bf1,Ng1,Rh1
+}
+% spell:on
+
+
+\vfill
+\pagebreak
+
+
+
+
+
+% Sherlock, little exercise 2
+\problem{Kidnapping}
+\difficulty{4}{5}
+
+On which square was the White queen captured?. \par
+
+% spell:off
+\manyboards{
+	ra8,qd8,ke8,ng8,rh8,
+	pa7,pb7,pc7,pe7,pf7,ph7,
+	nc6,pe6,ph6,
+	Pb3,
+	Na2,Pb2,Pc2,Pd2,Pe2,Pf2,Pg2,Ph2,
+	Ra1,Ke1,Rh1
+}
+% spell:on
+
+\begin{solution}
+	White is missing a queen, both bishops, and one knight. \par
+	The black pawns on E6 and H6 account for two captures.
+
+	\vspace{2mm}
+
+	Neither white bishop could've been captured by these pawns,
+	since both are trapped by their pawns. Thus, these black pawns must have captured a queen and a knight.
+
+	\vspace{4mm}
+
+	The white pawn on B3 must have captured a black bishop. \par
+	The white queen got onto the board through A2. \par
+	Therefore, the pawn on B3 made its capture before the queen escaped,
+	and the black bishop was captured before the white queen.
+
+	\vspace{4mm}
+
+	Similarly, the bishop from C8 must have been
+	captured on B3 after the capture on E6, since it
+	got on the board through D7.
+
+
+	\vspace{4mm}
+
+	The capture on E6 was made before the capture on B3 (black bishop),
+	which was made before the white queen was captured.
+	Therefore, the white queen was not captured on E6, and must
+	have been lost on H6.
+\end{solution}
+
+\vfill
+\pagebreak
+
 
 
 
 % Arabian Knights 4
 \problem{A missing piece}
-\difficulty{4}{5}
+\difficulty{6}{8}
 
 
 There is a piece at G4, marked with a $\odot$. \par
@@ -251,7 +391,7 @@ 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 accouted 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 accounted 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/Advanced/Retrograde Analysis/parts/04 hard.tex

@@ -2,7 +2,7 @@
 
 % Arabian Knights 5
 \problem{The hidden castle}
-\difficulty{7}{7}
+\difficulty{8}{8}
 
 There is a white castle hidden on this board. Where is it? \par
 None of the royalty has moved or been under attack. \par
@@ -30,7 +30,7 @@ None of the royalty has moved or been under attack. \par
 
 % Arabian Knights 6
 \problem{Who moved last?}
-\difficulty{7}{7}
+\difficulty{8}{8}
 
 After many moves of chess, the board looks as follows. \par
 Who moved last? \par
@@ -58,7 +58,7 @@ Who moved last? \par
 
 % Arabian Knights 3
 \problem{The king in disguise}<kingdisguise>
-\difficulty{7}{7}
+\difficulty{8}{8}
 
 The white king is exploring his kingdom under a disguise. He could look like any piece of any color.\par
 Show that he must be on C7.
@@ -119,7 +119,7 @@ Show that he must be on C7.
 
 % Arabian Knights 3
 \problem{The king in disguise once more}
-\difficulty{5}{7}
+\difficulty{5}{8}
 
 The white king is again exploring his kingdom, now under a different disguise. Where is he? \par
 \hint{\say{different disguise} implies that the white king looks like a different piece!}

src/Intermediate/Instant Insanity/main.tex

@@ -331,7 +331,7 @@
 	representing all four cubes. \\
 
 	\begin{center} \begin{small}
-	\begin{tikzpicture} \label{pic:II_comfiguration}
+	\begin{tikzpicture} \label{pic:II_configuration}
 		\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

@@ -0,0 +1,11 @@
+#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

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

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

@@ -0,0 +1,50 @@
+#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

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

src/Warm-Ups/Painting/main.typ

@@ -2,9 +2,8 @@
 #import "@preview/cetz:0.4.2"
 
 #show: handout.with(
-  title: [Warm-Up: What's an AST?],
+  title: [Warm-Up: The Painting],
   by: "Mark",
-  subtitle: "Based on a true story.",
 )
 
 #problem()
@@ -13,7 +12,8 @@ Hang the painting on two nails so that if either is removed, the painting falls.
 
 #v(2mm)
 You may detach the string as you hang the painting, but it must be re-attached once you're done. \
-#hint[The solution to this problem isn't a "think outside the box" trick, it's a clever wrapping of the string.]
+
+The solution to this problem isn't a trick, it's a clever wrapping of the string.
 
 
 #v(2mm)