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/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)
@@ -373,3 +333,121 @@ Which bishop was it, and what did it capture? \par
 
 \vfill
 \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

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/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/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)