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2
Advanced/COM/parts/1 continuous.tex
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@ -44,7 +44,7 @@ you can get it to balance on the beveled edge, as seen in Figure
\end{figure} \end{figure}
\problem{} \problem{}
See Figure \ref{soda filled}. Let's take the can to be massless and intially empty. Let's also assume that we live in two dimensions. We start slowly filling it up with soda to a vertical height $h$. What is $h$ just before the can tips over? See Figure \ref{soda filled}. Let's take the can to be massless and initially empty. Let's also assume that we live in two dimensions. We start slowly filling it up with soda to a vertical height $h$. What is $h$ just before the can tips over?
\vfill \vfill

2
Advanced/Compression/parts/2 lzss.tex
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@ -35,7 +35,9 @@ Then, decode the following:
\begin{solution} \begin{solution}
% spell:off
\texttt{ABCD$\cdot$ABCD$\cdot$BABABA$\cdot$ABCD$\cdot$ABCD} becomes \texttt{[ABCD<4, 4> BA<2,4> ABCD<4,4>]}. \texttt{ABCD$\cdot$ABCD$\cdot$BABABA$\cdot$ABCD$\cdot$ABCD} becomes \texttt{[ABCD<4, 4> BA<2,4> ABCD<4,4>]}.
% spell:on
\linehack{} \linehack{}

3
Advanced/Compression/parts/3 huffman.tex
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@ -256,7 +256,8 @@ Now, do the opposite: draw a tree that encodes \texttt{DEACBDD} \textit{less} ef
\remark{} \remark{}
As we just saw, constructing a prefix-free code is fairly easy. \par As we just saw, constructing a prefix-free code is fairly easy. \par
Constucting the \textit{most efficient} prefix-free code for a given message is a bit more difficult. \par Constructing the \textit{most efficient} prefix-free code for a
given message is a bit more difficult. \par
\pagebreak \pagebreak

2
Advanced/Cryptography/parts/4 DiffieHellman.tex
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@ -101,7 +101,7 @@ Run the algorithm. What is the resulting shared secret?
\begin{solution} \begin{solution}
$g^b = 5$\par $g^b = 5$\par
$g^a = 6$\par $g^a = 6$\par
$g^{ab} = g^{ba} = 9$ $g^{ab} = g^{ba} = 9$ % spell:disable-line
\end{solution} \end{solution}

8
Advanced/De Bruijn/parts/4 sturmian.tex
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@ -276,7 +276,7 @@ Attempt the above construction a few times. Is $w$ a minimal Sturmian word?
\theorem{}<sturmanthm> \theorem{}<sturmanthm>
We can construct a miminal Sturmian word of order $n \geq 3$ as follows: We can construct a minimal Sturmian word of order $n \geq 3$ as follows:
\begin{itemize} \begin{itemize}
\item Start with $G_2$, create $R_2$ by removing one edge. \item Start with $G_2$, create $R_2$ by removing one edge.
\item Construct $\mathcal{L}(G_2)$, remove an edge if necessary. \par \item Construct $\mathcal{L}(G_2)$, remove an edge if necessary. \par
@ -315,7 +315,7 @@ Construct a minimal Sturmain word of order 4.
$R_4 = \mathcal{L}(R_3)$ is then as shown below, producing the $R_4 = \mathcal{L}(R_3)$ is then as shown below, producing the
order $4$ minimal Sturman word \texttt{11110000}. Disconnected order $4$ minimal Sturman word \texttt{11110000}. Disconnected
nodes are ommited. nodes are omitted.
\begin{center} \begin{center}
\begin{tikzpicture} \begin{tikzpicture}
@ -345,7 +345,7 @@ Construct a minimal Sturmain word of order 5.
\begin{solution} \begin{solution}
Use $R_4$ from \ref{sturmianfour} to construct $R_5$, shown below. \par Use $R_4$ from \ref{sturmianfour} to construct $R_5$, shown below. \par
Disconnected nodes are ommited. Disconnected nodes are omitted.
\begin{center} \begin{center}
\begin{tikzpicture} \begin{tikzpicture}
@ -375,7 +375,7 @@ Construct a minimal Sturmain word of order 5.
\problem{} \problem{}
Argue that the words we get by \ref{sturmanthm} are mimimal Sturmain words. \par Argue that the words we get by \ref{sturmanthm} are minimal Sturmain words. \par
That is, the word $w$ has length $2n$ and $\mathcal{S}_m(w) = m + 1$ for all $m \leq n$. That is, the word $w$ has length $2n$ and $\mathcal{S}_m(w) = m + 1$ for all $m \leq n$.
\begin{solution} \begin{solution}

4
Advanced/Esoteric Languages/parts/00 turing.tex
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@ -1,11 +1,11 @@
\section{Turing} \section{Turing}
\definition{} \definition{}
An \textit{esoteric programming langauge} is a programming langauge made for fun. \par An \textit{esoteric programming language} is a programming language made for fun. \par
We'll work with two such languages today: \textit{Turing} and \textit{Befunge}. We'll work with two such languages today: \textit{Turing} and \textit{Befunge}.
\definition{} \definition{}
\textit{Turing} is one of the most famous esoteric langauges, and is extremely minimalist. \par \textit{Turing} is one of the most famous esoteric languages, and is extremely minimalist. \par
It consists only of eight symbols, a data pointer, and an instruction pointer. It consists only of eight symbols, a data pointer, and an instruction pointer.
Turing's eight symbols are as follows: Turing's eight symbols are as follows:

2
Advanced/Esoteric Languages/parts/01 befunge.tex
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@ -1,7 +1,7 @@
\section{Befunge} \section{Befunge}
\definition{} \definition{}
\textit{Befunge} is another esoteric programming langauge, designed to be very difficult to compile. \par \textit{Befunge} is another esoteric programming language, designed to be very difficult to compile. \par
It consists of a \say{field} of instructions, a two-dimensional program counter, and a stack of values. \par It consists of a \say{field} of instructions, a two-dimensional program counter, and a stack of values. \par
\vspace{2mm} \vspace{2mm}

2
Advanced/Generating Functions/parts/02 dice.tex
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@ -79,7 +79,7 @@ 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 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 numer of ways the sum of nonstandard dice rolls as $k$ is equal to the number of ways the sum of
two standard dice rolls as $k$. two standard dice rolls as $k$.
\hint{factor polynomials.} \hint{factor polynomials.}
\begin{solution} \begin{solution}

2
Advanced/Geometric Optimization/main.tex
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@ -170,7 +170,7 @@
\problem{} \problem{}
Draw a circle, then draw two distinct tangents $\ell_1$ and $\ell_2$ that intersect at point $A$. \par Draw a circle, then draw two distinct tangents $\ell_1$ and $\ell_2$ that intersect at point $A$. \par
Let $P$ be a point on the circle between the tangents, and $BC$ be the tangent at that point. Let $P$ be a point on the circle between the tangents, and $BC$ be the tangent at that point.
Describe how $P$ shoud be selected in order to minimize the perimeter of triangle $ABC$. Describe how $P$ should be selected in order to minimize the perimeter of triangle $ABC$.
\begin{center} \begin{center}
\begin{tikzpicture}[scale = 2] \begin{tikzpicture}[scale = 2]

6
Advanced/Geometry of Masses/parts/1 continuous.tex
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@ -44,7 +44,11 @@ you can get it to balance on the beveled edge, as seen in Figure
\end{figure} \end{figure}
\problem{} \problem{}
See Figure \ref{soda filled}. Let's take the can to be massless and intially empty. Let's also assume that we live in two dimensions. We start slowly filling it up with soda to a vertical height $h$. What is $h$ just before the can tips over? See Figure \ref{soda filled}.
Let's take the can to be massless and initially empty.
Let's also assume that we live in two dimensions.
We start slowly filling it up with soda to a vertical height $h$.
What is $h$ just before the can tips over?
\vfill \vfill

2
Advanced/Intro to Proofs/main.tex
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@ -305,7 +305,7 @@
\problem{} \problem{}
Consider a rectangular chocolate bar of arbitrary size. \par Consider a rectangular chocolate bar of arbitrary size. \par
What is the minimum number of breaks you need to make to What is the minimum number of breaks you need to make to
seperate all its pieces? separate all its pieces?
\begin{solution} \begin{solution}
number of squares minus one, simple proof by induction. number of squares minus one, simple proof by induction.

2
Advanced/Introduction to Quantum/src/main.tex
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@ -32,7 +32,7 @@
\def\bra#1{\left\langle#1\right|} \def\bra#1{\left\langle#1\right|}
% TODO: spend more time on probabalistic bits. % TODO: spend more time on probabalastic bits.
% This could even be its own handout, especially % This could even be its own handout, especially
% for younger classes! % for younger classes!

2
Advanced/Introduction to Quantum/src/parts/01 bits.tex
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@ -498,7 +498,7 @@ Compute the following product:
\generic{Remark:} \generic{Remark:}
Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par
We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangably. We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably.
\pagebreak \pagebreak

2
Advanced/Introduction to Quantum/src/parts/02 qubit.tex
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@ -18,7 +18,7 @@ $\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one
\note[Note]{$\bra{0}$ is called a \say{bra,} but we won't worry about that for now.} \note[Note]{$\bra{0}$ is called a \say{bra,} but we won't worry about that for now.}
\vspace{2mm} \vspace{2mm}
This is very similiar to the \say{box} $[~]$ notation we used for probabilistic bits. \par This is very similar to the \say{box} $[~]$ notation we used for probabilistic bits. \par
As before, we will write $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ As before, we will write $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$. and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.

2
Advanced/Introduction to Quantum/src/parts/06 hxh.tex
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@ -64,7 +64,7 @@ and like $X$ otherwise. The two circuits above illustrate this fact---take a loo
\vspace{2mm} \vspace{2mm}
Of course, we can give a gate multiple controls. \par Of course, we can give a gate multiple controls. \par
An $X$ gate with multiplie controls behaves like an $X$ gate if... An $X$ gate with multiple controls behaves like an $X$ gate if...
\begin{itemize} \begin{itemize}
\item all non-inverted controls are $\ket{1}$, and \item all non-inverted controls are $\ket{1}$, and
\item all inverted controls are $\ket{0}$ \item all inverted controls are $\ket{0}$

5
Advanced/Introduction to Quantum/src/parts/08 teleport.tex
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@ -1,7 +1,8 @@
\section{Quantum Teleportation} \section{Quantum Teleportation}
Superdense coding lets us convert quantum bandwidth into classical bandwidth. \par Superdense coding lets us convert quantum bandwidth into classical bandwidth. \par
Quantum teleporation does the opposite, using two classical bits and an entangled pair to transmit a quantum state. Quantum teleportation does the opposite, using two classical bits and an entangled pair
to transmit a quantum state.
\generic{Setup:} \generic{Setup:}
Again, suppose Alice and Bob each have half of a $\ket{\Phi^+}$ state. \par Again, suppose Alice and Bob each have half of a $\ket{\Phi^+}$ state. \par
@ -131,7 +132,7 @@ With an informal proof, show that it is not possible to use superdense coding to
more than two classical bits through an entangled two-qubit quantum state. more than two classical bits through an entangled two-qubit quantum state.
\begin{solution} \begin{solution}
If superdense coding was any more efficient, we could repeatedly apply superdense coding and quantum teleporation, If superdense coding was any more efficient, we could repeatedly apply superdense coding and quantum teleportation,
to compress an arbitrary number of bits into two \say{seed} bits. to compress an arbitrary number of bits into two \say{seed} bits.
\linehack{} \linehack{}

6
Advanced/Nonstandard Analysis/parts/extensions.tex
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@ -52,7 +52,7 @@ An ordered field must satisfy the following properties:
An ordered field that contains $\mathbb{R}$ is called an \textit{extension} of $\mathbb{R}$. An ordered field that contains $\mathbb{R}$ is called an \textit{extension} of $\mathbb{R}$.
\definition{} \definition{}
The \textit{Archimedian property} states the following: \par The \textit{Archimedean property} states the following: \par
For all positive $x, y$, there exists an $n$ so that $nx \geq y$. For all positive $x, y$, there exists an $n$ so that $nx \geq y$.
\theorem{} \theorem{}
@ -149,7 +149,9 @@ In an ordered field, the \textit{magnitude} of a number x is defined as follows:
\end{equation*} \end{equation*}
\definition{} \definition{}
We say an element $\delta$ of an ordered field is \textit{infinitesimal} if $|nd| < 1$ for all $n \in \mathbb{Z^+}$. \par We say an element $\delta$ of an ordered field is \textit{infinitesimal} if
$|nd| < 1$ % spell:disable-line
for all $n \in \mathbb{Z^+}$. \par
\note{Note that $\mathbb{Z}^+$ is a subset of any nonarchimedian extension of $\mathbb{R}$.} \par \note{Note that $\mathbb{Z}^+$ is a subset of any nonarchimedian extension of $\mathbb{R}$.} \par
\vspace{2mm} \vspace{2mm}

5
Advanced/Pidgeonhole Problems/main.tex
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@ -380,7 +380,10 @@
In a year, there are $364$ clean breaks. This leaves $499 - 364 = 135$ ``dirty'' breaks. \par In a year, there are $364$ clean breaks. This leaves $499 - 364 = 135$ ``dirty'' breaks. \par
We therefore have $135$ places to start a block on a dirty break, and $135$ places to end a block on a dirty break. This gives us a maximum of $270$ dirty blocks. \par We therefore have $135$ places to start a block on a dirty break, and $135$ places to end a block on a dirty break. This gives us a maximum of $270$ dirty blocks. \par
However, there are $401$ possible blocks, since we can start one at the $1^{\text{st}}, 2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par % spell:off
However, there are $401$ possible blocks,
since we can start one at the $1^{\text{st}},2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par
% spell:on
Out of $401$ blocks, a maximum of $270$ can be dirty. We are therefore guaranteed at least $131$ clean blocks. This completes the problem---each clean block represents a set of consecutive, whole days during which exactly 100 espressos were consumed. Out of $401$ blocks, a maximum of $270$ can be dirty. We are therefore guaranteed at least $131$ clean blocks. This completes the problem---each clean block represents a set of consecutive, whole days during which exactly 100 espressos were consumed.

14
Advanced/Retrograde Analysis/parts/02 easy.tex
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@ -12,11 +12,13 @@ What was Black's last move, and what was White's last move? \par
There a few empty boards at the end of this handout as well. There a few empty boards at the end of this handout as well.
} }
% spell:off
\manyboards{ \manyboards{
ka8,Kc8, ka8,Kc8,
Ph2, Ph2,
Bg1 Bg1
} }
% spell:on
\begin{solution} \begin{solution}
It's pretty clear that Black just moved out of check from A7. It's pretty clear that Black just moved out of check from A7.
@ -59,12 +61,14 @@ In the game below, no pieces have moved from a black square to a white square, o
There is a pawn at G3. What color is it? \par There is a pawn at G3. What color is it? \par
As before, White started on the bottom of the board. As before, White started on the bottom of the board.
% spell:off
\manyboards{ \manyboards{
ke8, ke8,
Kb4, Kb4,
Ug3, Ug3,
Pd2,Pf2 Pd2,Pf2
} }
% spell:on
\begin{solution} \begin{solution}
The white king is the key to this solution. How did it get off of E1? \par The white king is the key to this solution. How did it get off of E1? \par
@ -99,11 +103,13 @@ As before, White started on the bottom of the board.
The black king has turned himself invisible. Unfortunately, his position is hopeless. \par The black king has turned himself invisible. Unfortunately, his position is hopeless. \par
Mate the king in one move. \par Mate the king in one move. \par
% spell:off
\manyboards{ \manyboards{
Ra8,rb8,Kf8, Ra8,rb8,Kf8,
Nb7,Pc7, Nb7,Pc7,
Pa6,Rc6 Pa6,Rc6
} }
% spell:on
\begin{solution} \begin{solution}
Since it is White's move, Black cannot be in check. \par Since it is White's move, Black cannot be in check. \par
@ -132,11 +138,13 @@ Mate the king in one move. \par
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. 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.
There is one more piece on the board, which isn't shown. What color square does it stand on? \par There is one more piece on the board, which isn't shown. What color square does it stand on? \par
% spell:off
\manyboards{ \manyboards{
ke8, ke8,
Pd2,Pf2, Pd2,Pf2,
Ke1 Ke1
} }
% spell:on
\begin{solution} \begin{solution}
@ -161,11 +169,13 @@ In the game below, no pieces have moved from a black square to a white square, o
The white king has made less than fourteen moves. \par The white king has made less than fourteen moves. \par
Use this information to show that a pawn was promoted. \par Use this information to show that a pawn was promoted. \par
% spell:off
\manyboards{ \manyboards{
ke8, ke8,
Pb2,Pd2, Pb2,Pd2,
Ke1 Ke1
} }
% spell:on
\begin{solution} \begin{solution}
Knights always move to a different colored square, so all four missing knights must have been captured on their home square. Knights always move to a different colored square, so all four missing knights must have been captured on their home square.
@ -196,10 +206,12 @@ Use this information to show that a pawn was promoted. \par
It is White's move. Have there been any promotions this game? \par It is White's move. Have there been any promotions this game? \par
% spell:off
\manyboards{ \manyboards{
Pb2,Pe2,kf2,Pg2,Ph2, Pb2,Pe2,kf2,Pg2,Ph2,
Bc1,Kd1,Rh1 Bc1,Kd1,Rh1
} }
% spell:on
\begin{solution} \begin{solution}
@ -231,6 +243,7 @@ It is White's move. Have there been any promotions this game? \par
It is Black's move. Can Black castle? \par It is Black's move. Can Black castle? \par
\hint{Remember the rules of chess: you may not castle if you've moved your rook.} \hint{Remember the rules of chess: you may not castle if you've moved your rook.}
% spell:off
\manyboards{ \manyboards{
ra8,bc8,ke8,rh8, ra8,bc8,ke8,rh8,
pa7,pc7,pe7,pg7, pa7,pc7,pe7,pg7,
@ -239,6 +252,7 @@ It is Black's move. Can Black castle? \par
Pb2,Pc2,Pd2,Pe2,Pf2,Pg2,Ph2, Pb2,Pc2,Pd2,Pe2,Pf2,Pg2,Ph2,
Bc1,Qd1,Ke1,Bf1 Bc1,Qd1,Ke1,Bf1
} }
% spell:on
\begin{solution} \begin{solution}
White's last move was with the pawn. \par White's last move was with the pawn. \par

11
Advanced/Retrograde Analysis/parts/03 medium.tex
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@ -7,6 +7,8 @@
The results of a game of chess are shown below. \par The results of a game of chess are shown below. \par
Did White start on the north or south side of the board? \par Did White start on the north or south side of the board? \par
% spell:off
\manyboards{ \manyboards{
ka8,Kc8, ka8,Kc8,
Qe7, Qe7,
@ -15,6 +17,7 @@ Did White start on the north or south side of the board? \par
Ph3, Ph3,
Bh1 Bh1
} }
% spell:on
\begin{solution} \begin{solution}
Let us first find White's last move. It wasn't with the pawns on D4 and E5, since Black wouldn't have a move before that. Let us first find White's last move. It wasn't with the pawns on D4 and E5, since Black wouldn't have a move before that.
@ -52,11 +55,13 @@ Did White start on the north or south side of the board? \par
The white king has again become invisible. Find him. \par The white king has again become invisible. Find him. \par
\hint{White started on the bottom. En passant.} \par \hint{White started on the bottom. En passant.} \par
% spell:off
\manyboards{ \manyboards{
rb5,bd5, rb5,bd5,
Ba4, Ba4,
kd1 kd1
} }
% spell:on
\begin{solution} \begin{solution}
@ -77,6 +82,7 @@ The white king has again become invisible. Find him. \par
\begin{minipage}{0.5\linewidth} \begin{minipage}{0.5\linewidth}
\begin{center} \begin{center}
% spell:off
\chessboard[ \chessboard[
setpieces = { setpieces = {
rb5, rb5,
@ -86,6 +92,7 @@ The white king has again become invisible. Find him. \par
kd1 kd1
} }
] ]
% spell:on
\end{center} \end{center}
\end{minipage} \end{minipage}
\hfill \hfill
@ -123,11 +130,13 @@ The white king has again become invisible. Find him. \par
White to move. Which side of the board did each color start on? \par White to move. Which side of the board did each color start on? \par
\hint{What was Black's last move? } \hint{What was Black's last move? }
% spell:off
\manyboards{ \manyboards{
Re3, Re3,
Nc2,Rd2, Nc2,Rd2,
Nd1,kf1,Kh1 Nd1,kf1,Kh1
} }
% spell:on
\begin{solution} \begin{solution}
Black's last move was from F2, where his king was in double-check from both a rook and a knight. Black's last move was from F2, where his king was in double-check from both a rook and a knight.
@ -175,6 +184,7 @@ White to move. Which side of the board did each color start on? \par
There is a piece at G4, marked with a $\odot$. \par There is a piece at G4, marked with a $\odot$. \par
What is it, and what is its color? \par What is it, and what is its color? \par
% spell:off
\manyboards{ \manyboards{
ra8,ke8,rh8, ra8,ke8,rh8,
pc7,pd7, pc7,pd7,
@ -185,6 +195,7 @@ What is it, and what is its color? \par
ba2,Pb2,Pc2,Pd2,Pf2,qg2,bh2, ba2,Pb2,Pc2,Pd2,Pf2,qg2,bh2,
Kc1,Rd1,nf1,Bh1 Kc1,Rd1,nf1,Bh1
} }
% spell:on
\makeatletter \makeatletter

8
Advanced/Retrograde Analysis/parts/04 hard.tex
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@ -7,6 +7,7 @@
There is a white castle hidden on this board. Where is it? \par There is a white castle hidden on this board. Where is it? \par
None of the royalty has moved or been under attack. \par None of the royalty has moved or been under attack. \par
% spell:off
\manyboards{ \manyboards{
nb8,qd8,ke8,ng8,rh8, nb8,qd8,ke8,ng8,rh8,
pa7,pb7,pc7,pf7,pg7, pa7,pb7,pc7,pf7,pg7,
@ -16,6 +17,7 @@ None of the royalty has moved or been under attack. \par
Pb2,Pd2,Pf2,Pg2, Pb2,Pd2,Pf2,Pg2,
Qd1,Ke1 Qd1,Ke1
} }
% spell:on
\begin{solution} \begin{solution}
See \say{The Hidden Castle} in \textit{The Chess Mysteries of the Arabian Knights}. See \say{The Hidden Castle} in \textit{The Chess Mysteries of the Arabian Knights}.
@ -33,6 +35,7 @@ None of the royalty has moved or been under attack. \par
After many moves of chess, the board looks as follows. \par After many moves of chess, the board looks as follows. \par
Who moved last? \par Who moved last? \par
% spell:off
\manyboards{ \manyboards{
ka8,Kc8,bf8,rh8, ka8,Kc8,bf8,rh8,
pb7,pc7,pf7,pg7, pb7,pc7,pf7,pg7,
@ -41,6 +44,7 @@ Who moved last? \par
Pa2,Pb2,Pd2,Pg2,Ph2, Pa2,Pb2,Pd2,Pg2,Ph2,
Ra1,Nb1,Bc1,Qd1,Rh1 Ra1,Nb1,Bc1,Qd1,Rh1
} }
% spell:on
\begin{solution} \begin{solution}
See \say{A Vital Decision} in \textit{The Chess Mysteries of the Arabian Knights}. See \say{A Vital Decision} in \textit{The Chess Mysteries of the Arabian Knights}.
@ -59,6 +63,7 @@ Who moved last? \par
The white king is exploring his kingdom under a disguise. He could look like any piece of any color.\par 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. Show that he must be on C7.
% spell:off
\manyboards{ \manyboards{
qa8,nb8,be8,Qg8,kh8, qa8,nb8,be8,Qg8,kh8,
pa7,Pb7,pc7,Nd7,pe7,Pf7,ph7, pa7,Pb7,pc7,Nd7,pe7,Pf7,ph7,
@ -66,6 +71,7 @@ Show that he must be on C7.
ra5,pb5,Rd5,Ph5, ra5,pb5,Rd5,Ph5,
Pa4,Nc4,Pe4,Bg4 Pa4,Nc4,Pe4,Bg4
} }
% spell:on
\begin{solution} \begin{solution}
Black is in check, so we know that it is Black's move and White is not in check.\par Black is in check, so we know that it is Black's move and White is not in check.\par
@ -118,6 +124,7 @@ Show that he must be on C7.
The white king is again exploring his kingdom, now under a different disguise. Where is he? \par 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!} \hint{\say{different disguise} implies that the white king looks like a different piece!}
% spell:off
\manyboards{ \manyboards{
nb8,be8,Qg8,kh8, nb8,be8,Qg8,kh8,
pa7,Pb7,pc7,Nd7,pe7,Pf7,ph7, pa7,Pb7,pc7,Nd7,pe7,Pf7,ph7,
@ -125,6 +132,7 @@ The white king is again exploring his kingdom, now under a different disguise. W
ra5,pb5,Rd5,Ph5, ra5,pb5,Rd5,Ph5,
Pa4,Nc4,Pe4,Bg4 Pa4,Nc4,Pe4,Bg4
} }
% spell:on
\begin{solution} \begin{solution}
Use the same arguments as before, but now assume that the king isn't a black pawn. Use the same arguments as before, but now assume that the king isn't a black pawn.

16
Advanced/Stopping Problems/parts/0 probability.tex
View File

@ -46,17 +46,17 @@ a \textit{random variable} is a function from $\Omega$ to a specified output set
For example, given the three-coin-toss sample space For example, given the three-coin-toss sample space
$\Omega = \{ $\Omega = \{
\texttt{TTT},~ \texttt{TTH},~ \texttt{THT},~ \texttt{TTT},~ \texttt{TTH},~ \texttt{THT},~ % spell:disable-line
\texttt{THH},~ \texttt{HTT},~ \texttt{HTH},~ \texttt{THH},~ \texttt{HTT},~ \texttt{HTH},~ % spell:disable-line
\texttt{HHT},~ \texttt{HHH} \texttt{HHT},~ \texttt{HHH} % spell:disable-line
\}$, \}$,
We can define a random variable $\mathcal{H}$ as \say{the number of heads in a throw of three coins}. \par We can define a random variable $\mathcal{H}$ as \say{the number of heads in a throw of three coins}. \par
As a function, $\mathcal{H}$ maps values in $\Omega$ to values in $\mathbb{Z}^+_0$ and is defined as: As a function, $\mathcal{H}$ maps values in $\Omega$ to values in $\mathbb{Z}^+_0$ and is defined as:
\begin{itemize} \begin{itemize}
\item $\mathcal{H}(\texttt{TTT}) = 0$ \item $\mathcal{H}(\texttt{TTT}) = 0$ % spell:disable-line
\item $\mathcal{H}(\texttt{TTH}) = 1$ \item $\mathcal{H}(\texttt{TTH}) = 1$ % spell:disable-line
\item $\mathcal{H}(\texttt{THT}) = 1$ \item $\mathcal{H}(\texttt{THT}) = 1$ % spell:disable-line
\item $\mathcal{H}(\texttt{THH}) = 2$ \item $\mathcal{H}(\texttt{THH}) = 2$ % spell:disable-line
\item ...and so on. \item ...and so on.
\end{itemize} \end{itemize}
@ -70,7 +70,7 @@ the set of outcomes that produce that value. \par
\vspace{2mm} \vspace{2mm}
For example, if we wanted to compute $\mathcal{P}(\mathcal{H} = 2)$, we would find For example, if we wanted to compute $\mathcal{P}(\mathcal{H} = 2)$, we would find
$\mathcal{P}\bigl(\{\texttt{THH}, \texttt{HTH}, \texttt{HHT}\}\bigr)$. $\mathcal{P}\bigl(\{\texttt{THH}, \texttt{HTH}, \texttt{HHT}\}\bigr)$. % spell:disable-line
\problem{} \problem{}

2
Advanced/Stopping Problems/parts/2 secretary.tex
View File

@ -191,7 +191,7 @@ what is the probability that we select the best candidate? \par
Call this probability $\phi_n(k)$. Call this probability $\phi_n(k)$.
\begin{solution} \begin{solution}
Using \ref{seca} and \ref{secb}, this is straightfoward: Using \ref{seca} and \ref{secb}, this is straightforward:
\[ \[
\phi_n(k) \phi_n(k)
= \sum_{x = k}^{n}\left( \frac{1}{n} \times \frac{k-1}{x-1} \right) = \sum_{x = k}^{n}\left( \frac{1}{n} \times \frac{k-1}{x-1} \right)

26
Intermediate/Instant Insanity/main.tex
View File

@ -120,7 +120,7 @@
\node[text = white] at (0 + 7.5, 0 + 1.5) {\textbf{Red}}; \node[text = white] at (0 + 7.5, 0 + 1.5) {\textbf{Red}};
\node[text = white] at (0 + 10.5, 0 + 1.5) {\textbf{Blue}}; \node[text = white] at (0 + 10.5, 0 + 1.5) {\textbf{Blue}};
\node[text = black] at (0 + 7.5, 0 + 4.5) {\textbf{Grn}}; \node[text = black] at (0 + 7.5, 0 + 4.5) {\textbf{Grn}};
\node[text = black] at (0 + 7.5, 0 - 1.5) {\textbf{Wht}}; \node[text = black] at (0 + 7.5, 0 - 1.5) {\textbf{Wht}}; % spell:disable-line
\filldraw [red] (21,0) -- (27,0) -- (27,3) -- \filldraw [red] (21,0) -- (27,0) -- (27,3) --
(21,3) -- (21,0); (21,3) -- (21,0);
@ -138,8 +138,8 @@
\node[text = white] at (21 + 1.5, 0 + 1.5) {\textbf{Red}}; \node[text = white] at (21 + 1.5, 0 + 1.5) {\textbf{Red}};
\node[text = white] at (21 + 4.5, 0 + 1.5) {\textbf{Red}}; \node[text = white] at (21 + 4.5, 0 + 1.5) {\textbf{Red}};
\node[text = black] at (21 + 7.5, 0 + 1.5) {\textbf{Grn}}; \node[text = black] at (21 + 7.5, 0 + 1.5) {\textbf{Grn}};
\node[text = black] at (21 + 10.5, 0 + 1.5) {\textbf{Wht}}; \node[text = black] at (21 + 10.5, 0 + 1.5) {\textbf{Wht}}; % spell:disable-line
\node[text = black] at (21 + 7.5, 0 + 4.5) {\textbf{Wht}}; \node[text = black] at (21 + 7.5, 0 + 4.5) {\textbf{Wht}}; % spell:disable-line
\node[text = white] at (21 + 7.5, 0 - 1.5) {\textbf{Blue}}; \node[text = white] at (21 + 7.5, 0 - 1.5) {\textbf{Blue}};
\filldraw [red] (0,-15) -- (3,-15) -- (3,-12) \filldraw [red] (0,-15) -- (3,-15) -- (3,-12)
@ -158,8 +158,8 @@
\node[text = white] at (0 + 1.5, -15 + 1.5) {\textbf{Red}}; \node[text = white] at (0 + 1.5, -15 + 1.5) {\textbf{Red}};
\node[text = black] at (0 + 4.5, -15 + 1.5) {\textbf{Grn}}; \node[text = black] at (0 + 4.5, -15 + 1.5) {\textbf{Grn}};
\node[text = white] at (0 + 7.5, -15 + 1.5) {\textbf{Blue}}; \node[text = white] at (0 + 7.5, -15 + 1.5) {\textbf{Blue}};
\node[text = black] at (0 + 10.5, -15 + 1.5) {\textbf{Wht}}; \node[text = black] at (0 + 10.5, -15 + 1.5) {\textbf{Wht}}; % spell:disable-line
\node[text = black] at (0 + 7.5, -15 + 4.5) {\textbf{Wht}}; \node[text = black] at (0 + 7.5, -15 + 4.5) {\textbf{Wht}}; % spell:disable-line
\node[text = white] at (0 + 7.5, -15 - 1.5) {\textbf{Blue}}; \node[text = white] at (0 + 7.5, -15 - 1.5) {\textbf{Blue}};
\filldraw [red] (21,-15) -- (24,-15) -- \filldraw [red] (21,-15) -- (24,-15) --
@ -181,7 +181,7 @@
\node[text = white] at (21 + 4.5, -15 + 1.5) {\textbf{Blue}}; \node[text = white] at (21 + 4.5, -15 + 1.5) {\textbf{Blue}};
\node[text = black] at (21 + 7.5, -15 + 1.5) {\textbf{Grn}}; \node[text = black] at (21 + 7.5, -15 + 1.5) {\textbf{Grn}};
\node[text = white] at (21 + 10.5, -15 + 1.5) {\textbf{Blue}}; \node[text = white] at (21 + 10.5, -15 + 1.5) {\textbf{Blue}};
\node[text = black] at (21 + 7.5, -15 + 4.5) {\textbf{Wht}}; \node[text = black] at (21 + 7.5, -15 + 4.5) {\textbf{Wht}}; % spell:disable-line
\node[text = black] at (21 + 7.5, -15 - 1.5) {\textbf{Grn}}; \node[text = black] at (21 + 7.5, -15 - 1.5) {\textbf{Grn}};
\end{tikzpicture} \end{tikzpicture}
@ -270,7 +270,7 @@
\node[text = white] at (0.5, 5.5) {\textbf{Blue}}; \node[text = white] at (0.5, 5.5) {\textbf{Blue}};
\node[text = white] at (0.5, 0.5) {\textbf{Red}}; \node[text = white] at (0.5, 0.5) {\textbf{Red}};
\node[text = black] at (5.5, 0.5) {\textbf{Wht}}; \node[text = black] at (5.5, 0.5) {\textbf{Wht}}; % spell:disable-line
\node[text = black] at (5.5, 5.5) {\textbf{Grn}}; \node[text = black] at (5.5, 5.5) {\textbf{Grn}};
\end{tikzpicture} \end{tikzpicture}
@ -321,7 +321,7 @@
\node[text = white] at (0.5, 5.5) {\textbf{Blue}}; \node[text = white] at (0.5, 5.5) {\textbf{Blue}};
\node[text = white] at (0.5, 0.5) {\textbf{Red}}; \node[text = white] at (0.5, 0.5) {\textbf{Red}};
\node[text = black] at (5.5, 0.5) {\textbf{Wht}}; \node[text = black] at (5.5, 0.5) {\textbf{Wht}}; % spell:disable-line
\node[text = black] at (5.5, 5.5) {\textbf{Grn}}; \node[text = black] at (5.5, 5.5) {\textbf{Grn}};
\end{tikzpicture} \end{tikzpicture}
\end{small} \end{center} \end{small} \end{center}
@ -400,7 +400,7 @@
\node[text = white] at (0.5, 5.5) {\textbf{Blue}}; \node[text = white] at (0.5, 5.5) {\textbf{Blue}};
\node[text = white] at (0.5, 0.5) {\textbf{Red}}; \node[text = white] at (0.5, 0.5) {\textbf{Red}};
\node[text = black] at (5.5, 0.5) {\textbf{Wht}}; \node[text = black] at (5.5, 0.5) {\textbf{Wht}}; % spell:disable-line
\node[text = black] at (5.5, 5.5) {\textbf{Grn}}; \node[text = black] at (5.5, 5.5) {\textbf{Grn}};
\end{tikzpicture} \end{tikzpicture}
\end{small} \end{center} \end{small} \end{center}
@ -508,7 +508,7 @@
\node[text = white] at (0.5, 5.5) {\textbf{Blue}}; \node[text = white] at (0.5, 5.5) {\textbf{Blue}};
\node[text = white] at (0.5, 0.5) {\textbf{Red}}; \node[text = white] at (0.5, 0.5) {\textbf{Red}};
\node[text = black] at (5.5, 0.5) {\textbf{Wht}}; \node[text = black] at (5.5, 0.5) {\textbf{Wht}}; % spell:disable-line
\node[text = black] at (5.5, 5.5) {\textbf{Grn}}; \node[text = black] at (5.5, 5.5) {\textbf{Grn}};
\end{tikzpicture} \end{tikzpicture}
\end{small} \end{center} \end{small} \end{center}
@ -611,7 +611,7 @@
\node[text = black] at (2.5, -1) {\textbf{Grn}}; \node[text = black] at (2.5, -1) {\textbf{Grn}};
\node[text = black] at (4.5, 0.5) {\textbf{Grn}}; \node[text = black] at (4.5, 0.5) {\textbf{Grn}};
\node[text = black] at (4.5, -1) {\textbf{Wht}}; \node[text = black] at (4.5, -1) {\textbf{Wht}}; % spell:disable-line
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
@ -657,10 +657,10 @@
\node[text = black] at (2.5, 0.5) {\textbf{Red}}; \node[text = black] at (2.5, 0.5) {\textbf{Red}};
\node[text = white] at (1.5, -1) {\textbf{Red}}; \node[text = white] at (1.5, -1) {\textbf{Red}};
\node[text = black] at (2.5, -1) {\textbf{Grn}}; \node[text = black] at (2.5, -1) {\textbf{Grn}};
\node[text = black] at (3.5, 0.5) {\textbf{Wht}}; \node[text = black] at (3.5, 0.5) {\textbf{Wht}}; % spell:disable-line
\node[text = white] at (3.5, -1) {\textbf{Blue}}; \node[text = white] at (3.5, -1) {\textbf{Blue}};
\node[text = black] at (4.5, 0.5) {\textbf{Grn}}; \node[text = black] at (4.5, 0.5) {\textbf{Grn}};
\node[text = black] at (4.5, -1) {\textbf{Wht}}; \node[text = black] at (4.5, -1) {\textbf{Wht}}; % spell:disable-line
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
\bigskip \bigskip

3
Intermediate/Vectors 1/main.tex
View File

@ -476,7 +476,10 @@
\section{Bonus} \section{Bonus}
\problem{Oldaque de Freitas' Puzzle} \problem{Oldaque de Freitas' Puzzle}
% spell:off
Two ladies are sitting in a street caf\'e, talking about their children. One lady says that she has three daughters. The product of the girls' ages equals 36 and the sum of their ages is the same as the number of the house across the street. The second lady replies that this information is not enough to figure out the age of each child. The first lady agrees and adds that her oldest daughter has beautiful blue eyes. The second lady then solves the puzzle. Please do the same. Two ladies are sitting in a street caf\'e, talking about their children. One lady says that she has three daughters. The product of the girls' ages equals 36 and the sum of their ages is the same as the number of the house across the street. The second lady replies that this information is not enough to figure out the age of each child. The first lady agrees and adds that her oldest daughter has beautiful blue eyes. The second lady then solves the puzzle. Please do the same.
% spell:on
\vfill \vfill

4
Misc/Warm-Ups/furniture.tex
View File

@ -39,7 +39,7 @@
\item Rosewood \item Rosewood
\end{itemize} \end{itemize}
The following contitions govern Irene's purchases: The following conditions govern Irene's purchases:
\begin{itemize} \begin{itemize}
\item Any vanity she buys is Maple. \item Any vanity she buys is Maple.
\item Any rosewood item she buys is a sideboard. \item Any rosewood item she buys is a sideboard.
@ -135,7 +135,7 @@
\problem{} \problem{}
Suppose the condition that Irene does not buy an oak table is Suppose the condition that Irene does not buy an oak table is
replaced with the condition that she does not buy a pine table. replaced with the condition that she does not buy a pine table.
If all the other contitions hold as originally given, which of the If all the other conditions hold as originally given, which of the
following cannot be true? following cannot be true?
\begin{itemize} \begin{itemize}
\item Irene buys an oak footstool. \item Irene buys an oak footstool.

4
Misc/Warm-Ups/mario.tex
View File

@ -40,7 +40,11 @@
A 11-way tie is possible, with a top score of 28: A 11-way tie is possible, with a top score of 28:
\begin{itemize} \begin{itemize}
\item Four players finish $1^\text{st}$, $3^\text{ed}$, $11^\text{th}$, and $12^\text{th}$; \item Four players finish $1^\text{st}$, $3^\text{ed}$, $11^\text{th}$, and $12^\text{th}$;
% spell:off
\item Four players finish $2^\text{nd}$, $4^\text{th}$, $9^\text{th}$, and $10^\text{th}$; \item Four players finish $2^\text{nd}$, $4^\text{th}$, $9^\text{th}$, and $10^\text{th}$;
% spell:on
\item Two players finish fifth twice and seventh twice, \item Two players finish fifth twice and seventh twice,
\item One player finishes sixth in each race. \item One player finishes sixth in each race.
\end{itemize} \end{itemize}

2
Misc/Warm-Ups/regex.tex
View File

@ -71,7 +71,9 @@
They specify exactly how many tokens to match: \par They specify exactly how many tokens to match: \par
\htexttt{ab\{2\}a} will match only \texttt{abba}. \par \htexttt{ab\{2\}a} will match only \texttt{abba}. \par
\htexttt{ab\{1,3\}a} will match only \texttt{aba}, \texttt{abba}, and \texttt{abbba}. \par \htexttt{ab\{1,3\}a} will match only \texttt{aba}, \texttt{abba}, and \texttt{abbba}. \par
% spell:off
\htexttt{ab\{2,\}a} will match any \texttt{ab...ba} with at least two \texttt{b}s. \htexttt{ab\{2,\}a} will match any \texttt{ab...ba} with at least two \texttt{b}s.
% spell:on
\vspace{5mm} \vspace{5mm}

4
resources/ormc_handout.cls
View File

@ -148,9 +148,9 @@
% Enum labels % Enum labels
\renewcommand\theenumi{\@arabic\c@enumi} \renewcommand\theenumi{\@arabic\c@enumi}
\renewcommand\theenumii{\@alph\c@enumii} \renewcommand\theenumii{\@alph\c@enumii} % spell:disable-line
\renewcommand\theenumiii{\@roman\c@enumiii} \renewcommand\theenumiii{\@roman\c@enumiii}
\renewcommand\theenumiv{\@Alph\c@enumiv} \renewcommand\theenumiv{\@Alph\c@enumiv} % spell:disable-line
\newcommand\labelenumi{\theenumi.} \newcommand\labelenumi{\theenumi.}
\newcommand\labelenumii{(\theenumii)} \newcommand\labelenumii{(\theenumii)}
\newcommand\labelenumiii{\theenumiii.} \newcommand\labelenumiii{\theenumiii.}

1
typos.toml
View File

@ -1,5 +1,6 @@
[default] [default]
extend-words."LSAT" = "LSAT" extend-words."LSAT" = "LSAT"
extend-words."ket" = "ket"
extend-ignore-re = [ extend-ignore-re = [
# spell:disable-line # spell:disable-line