diff --git a/Advanced/COM/parts/1 continuous.tex b/Advanced/COM/parts/1 continuous.tex index 591c1ba..6ee21a6 100644 --- a/Advanced/COM/parts/1 continuous.tex +++ b/Advanced/COM/parts/1 continuous.tex @@ -44,7 +44,7 @@ you can get it to balance on the beveled edge, as seen in Figure \end{figure} \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 diff --git a/Advanced/Compression/parts/2 lzss.tex b/Advanced/Compression/parts/2 lzss.tex index adf6e43..8ecae87 100644 --- a/Advanced/Compression/parts/2 lzss.tex +++ b/Advanced/Compression/parts/2 lzss.tex @@ -35,7 +35,9 @@ Then, decode the following: \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>]}. + % spell:on \linehack{} diff --git a/Advanced/Compression/parts/3 huffman.tex b/Advanced/Compression/parts/3 huffman.tex index 0a15c7e..3fab653 100644 --- a/Advanced/Compression/parts/3 huffman.tex +++ b/Advanced/Compression/parts/3 huffman.tex @@ -256,7 +256,8 @@ Now, do the opposite: draw a tree that encodes \texttt{DEACBDD} \textit{less} ef \remark{} 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 diff --git a/Advanced/Cryptography/parts/4 DiffieHellman.tex b/Advanced/Cryptography/parts/4 DiffieHellman.tex index 8a362d1..4044bf4 100755 --- a/Advanced/Cryptography/parts/4 DiffieHellman.tex +++ b/Advanced/Cryptography/parts/4 DiffieHellman.tex @@ -101,7 +101,7 @@ Run the algorithm. What is the resulting shared secret? \begin{solution} $g^b = 5$\par $g^a = 6$\par - $g^{ab} = g^{ba} = 9$ + $g^{ab} = g^{ba} = 9$ % spell:disable-line \end{solution} diff --git a/Advanced/De Bruijn/parts/4 sturmian.tex b/Advanced/De Bruijn/parts/4 sturmian.tex index 9458e74..97b87f0 100644 --- a/Advanced/De Bruijn/parts/4 sturmian.tex +++ b/Advanced/De Bruijn/parts/4 sturmian.tex @@ -276,7 +276,7 @@ Attempt the above construction a few times. Is $w$ a minimal Sturmian word? \theorem{} -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} \item Start with $G_2$, create $R_2$ by removing one edge. \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 order $4$ minimal Sturman word \texttt{11110000}. Disconnected - nodes are ommited. + nodes are omitted. \begin{center} \begin{tikzpicture} @@ -345,7 +345,7 @@ Construct a minimal Sturmain word of order 5. \begin{solution} Use $R_4$ from \ref{sturmianfour} to construct $R_5$, shown below. \par - Disconnected nodes are ommited. + Disconnected nodes are omitted. \begin{center} \begin{tikzpicture} @@ -375,7 +375,7 @@ Construct a minimal Sturmain word of order 5. \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 $m \leq n$. \begin{solution} diff --git a/Advanced/Esoteric Languages/parts/00 turing.tex b/Advanced/Esoteric Languages/parts/00 turing.tex index 089eeb1..0035db9 100755 --- a/Advanced/Esoteric Languages/parts/00 turing.tex +++ b/Advanced/Esoteric Languages/parts/00 turing.tex @@ -1,11 +1,11 @@ \section{Turing} \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}. \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. Turing's eight symbols are as follows: diff --git a/Advanced/Esoteric Languages/parts/01 befunge.tex b/Advanced/Esoteric Languages/parts/01 befunge.tex index f63bbd8..ecae011 100755 --- a/Advanced/Esoteric Languages/parts/01 befunge.tex +++ b/Advanced/Esoteric Languages/parts/01 befunge.tex @@ -1,7 +1,7 @@ \section{Befunge} \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 \vspace{2mm} diff --git a/Advanced/Generating Functions/parts/02 dice.tex b/Advanced/Generating Functions/parts/02 dice.tex index 92ee25b..f893350 100755 --- a/Advanced/Generating Functions/parts/02 dice.tex +++ b/Advanced/Generating Functions/parts/02 dice.tex @@ -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 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 $k$. \hint{factor polynomials.} \begin{solution} diff --git a/Advanced/Geometric Optimization/main.tex b/Advanced/Geometric Optimization/main.tex index e2b6d43..55e6fe3 100755 --- a/Advanced/Geometric Optimization/main.tex +++ b/Advanced/Geometric Optimization/main.tex @@ -170,7 +170,7 @@ \problem{} 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. - 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{tikzpicture}[scale = 2] diff --git a/Advanced/Geometry of Masses/parts/1 continuous.tex b/Advanced/Geometry of Masses/parts/1 continuous.tex index 591c1ba..ece665b 100644 --- a/Advanced/Geometry of Masses/parts/1 continuous.tex +++ b/Advanced/Geometry of Masses/parts/1 continuous.tex @@ -44,7 +44,11 @@ you can get it to balance on the beveled edge, as seen in Figure \end{figure} \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 diff --git a/Advanced/Intro to Proofs/main.tex b/Advanced/Intro to Proofs/main.tex index 54727a0..e19ab18 100755 --- a/Advanced/Intro to Proofs/main.tex +++ b/Advanced/Intro to Proofs/main.tex @@ -305,7 +305,7 @@ \problem{} Consider a rectangular chocolate bar of arbitrary size. \par What is the minimum number of breaks you need to make to - seperate all its pieces? + separate all its pieces? \begin{solution} number of squares minus one, simple proof by induction. diff --git a/Advanced/Introduction to Quantum/src/main.tex b/Advanced/Introduction to Quantum/src/main.tex index 2a964f2..97b8fbb 100755 --- a/Advanced/Introduction to Quantum/src/main.tex +++ b/Advanced/Introduction to Quantum/src/main.tex @@ -32,7 +32,7 @@ \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 % for younger classes! diff --git a/Advanced/Introduction to Quantum/src/parts/01 bits.tex b/Advanced/Introduction to Quantum/src/parts/01 bits.tex index ba444dc..6a98c52 100644 --- a/Advanced/Introduction to Quantum/src/parts/01 bits.tex +++ b/Advanced/Introduction to Quantum/src/parts/01 bits.tex @@ -498,7 +498,7 @@ Compute the following product: \generic{Remark:} 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 diff --git a/Advanced/Introduction to Quantum/src/parts/02 qubit.tex b/Advanced/Introduction to Quantum/src/parts/02 qubit.tex index 86839ed..f302e19 100644 --- a/Advanced/Introduction to Quantum/src/parts/02 qubit.tex +++ b/Advanced/Introduction to Quantum/src/parts/02 qubit.tex @@ -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.} \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 $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$. diff --git a/Advanced/Introduction to Quantum/src/parts/06 hxh.tex b/Advanced/Introduction to Quantum/src/parts/06 hxh.tex index cd99e92..ee3b0f9 100644 --- a/Advanced/Introduction to Quantum/src/parts/06 hxh.tex +++ b/Advanced/Introduction to Quantum/src/parts/06 hxh.tex @@ -64,7 +64,7 @@ and like $X$ otherwise. The two circuits above illustrate this fact---take a loo \vspace{2mm} 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} \item all non-inverted controls are $\ket{1}$, and \item all inverted controls are $\ket{0}$ diff --git a/Advanced/Introduction to Quantum/src/parts/08 teleport.tex b/Advanced/Introduction to Quantum/src/parts/08 teleport.tex index fd487e0..cc0082c 100644 --- a/Advanced/Introduction to Quantum/src/parts/08 teleport.tex +++ b/Advanced/Introduction to Quantum/src/parts/08 teleport.tex @@ -1,7 +1,8 @@ \section{Quantum Teleportation} 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:} 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. \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. \linehack{} diff --git a/Advanced/Nonstandard Analysis/parts/extensions.tex b/Advanced/Nonstandard Analysis/parts/extensions.tex index 90c1ecd..e0cb5d7 100644 --- a/Advanced/Nonstandard Analysis/parts/extensions.tex +++ b/Advanced/Nonstandard Analysis/parts/extensions.tex @@ -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 $\mathbb{R}$. \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 $nx \geq y$. \theorem{} @@ -149,7 +149,9 @@ In an ordered field, the \textit{magnitude} of a number x is defined as follows: \end{equation*} \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 \vspace{2mm} diff --git a/Advanced/Pidgeonhole Problems/main.tex b/Advanced/Pidgeonhole Problems/main.tex index 8d3f40b..2cc4279 100755 --- a/Advanced/Pidgeonhole Problems/main.tex +++ b/Advanced/Pidgeonhole Problems/main.tex @@ -380,7 +380,10 @@ 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 - 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. diff --git a/Advanced/Retrograde Analysis/parts/02 easy.tex b/Advanced/Retrograde Analysis/parts/02 easy.tex index d3f5263..c51140e 100644 --- a/Advanced/Retrograde Analysis/parts/02 easy.tex +++ b/Advanced/Retrograde Analysis/parts/02 easy.tex @@ -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. } +% spell:off \manyboards{ ka8,Kc8, Ph2, Bg1 } +% spell:on \begin{solution} 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 As before, White started on the bottom of the board. +% spell:off \manyboards{ ke8, Kb4, Ug3, Pd2,Pf2 } +% spell:on \begin{solution} 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 Mate the king in one move. \par +% spell:off \manyboards{ Ra8,rb8,Kf8, Nb7,Pc7, Pa6,Rc6 } +% spell:on \begin{solution} 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. There is one more piece on the board, which isn't shown. What color square does it stand on? \par +% spell:off \manyboards{ ke8, Pd2,Pf2, Ke1 } +% spell:on \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 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. @@ -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 +% spell:off \manyboards{ Pb2,Pe2,kf2,Pg2,Ph2, Bc1,Kd1,Rh1 } +% spell:on \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 \hint{Remember the rules of chess: you may not castle if you've moved your rook.} +% spell:off \manyboards{ ra8,bc8,ke8,rh8, pa7,pc7,pe7,pg7, @@ -239,6 +252,7 @@ It is Black's move. Can Black castle? \par Pb2,Pc2,Pd2,Pe2,Pf2,Pg2,Ph2, Bc1,Qd1,Ke1,Bf1 } +% spell:on \begin{solution} White's last move was with the pawn. \par diff --git a/Advanced/Retrograde Analysis/parts/03 medium.tex b/Advanced/Retrograde Analysis/parts/03 medium.tex index 3bf78ca..7ba70d8 100644 --- a/Advanced/Retrograde Analysis/parts/03 medium.tex +++ b/Advanced/Retrograde Analysis/parts/03 medium.tex @@ -7,6 +7,8 @@ The results of a game of chess are shown below. \par Did White start on the north or south side of the board? \par + +% spell:off \manyboards{ ka8,Kc8, Qe7, @@ -15,6 +17,7 @@ Did White start on the north or south side of the board? \par Ph3, Bh1 } +% spell:on \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. @@ -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 \hint{White started on the bottom. En passant.} \par +% spell:off \manyboards{ rb5,bd5, Ba4, kd1 } +% spell:on \begin{solution} @@ -77,6 +82,7 @@ The white king has again become invisible. Find him. \par \begin{minipage}{0.5\linewidth} \begin{center} + % spell:off \chessboard[ setpieces = { rb5, @@ -86,6 +92,7 @@ The white king has again become invisible. Find him. \par kd1 } ] + % spell:on \end{center} \end{minipage} \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 \hint{What was Black's last move? } +% spell:off \manyboards{ Re3, Nc2,Rd2, Nd1,kf1,Kh1 } +% spell:on \begin{solution} 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 What is it, and what is its color? \par +% spell:off \manyboards{ ra8,ke8,rh8, pc7,pd7, @@ -185,6 +195,7 @@ What is it, and what is its color? \par ba2,Pb2,Pc2,Pd2,Pf2,qg2,bh2, Kc1,Rd1,nf1,Bh1 } +% spell:on \makeatletter diff --git a/Advanced/Retrograde Analysis/parts/04 hard.tex b/Advanced/Retrograde Analysis/parts/04 hard.tex index f8c2160..4ca8eb1 100644 --- a/Advanced/Retrograde Analysis/parts/04 hard.tex +++ b/Advanced/Retrograde Analysis/parts/04 hard.tex @@ -7,6 +7,7 @@ There is a white castle hidden on this board. Where is it? \par None of the royalty has moved or been under attack. \par +% spell:off \manyboards{ nb8,qd8,ke8,ng8,rh8, pa7,pb7,pc7,pf7,pg7, @@ -16,6 +17,7 @@ None of the royalty has moved or been under attack. \par Pb2,Pd2,Pf2,Pg2, Qd1,Ke1 } +% spell:on \begin{solution} 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 Who moved last? \par +% spell:off \manyboards{ ka8,Kc8,bf8,rh8, pb7,pc7,pf7,pg7, @@ -41,6 +44,7 @@ Who moved last? \par Pa2,Pb2,Pd2,Pg2,Ph2, Ra1,Nb1,Bc1,Qd1,Rh1 } +% spell:on \begin{solution} 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 Show that he must be on C7. +% spell:off \manyboards{ qa8,nb8,be8,Qg8,kh8, pa7,Pb7,pc7,Nd7,pe7,Pf7,ph7, @@ -66,6 +71,7 @@ Show that he must be on C7. ra5,pb5,Rd5,Ph5, Pa4,Nc4,Pe4,Bg4 } +% spell:on \begin{solution} 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 \hint{\say{different disguise} implies that the white king looks like a different piece!} +% spell:off \manyboards{ nb8,be8,Qg8,kh8, 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, Pa4,Nc4,Pe4,Bg4 } +% spell:on \begin{solution} Use the same arguments as before, but now assume that the king isn't a black pawn. diff --git a/Advanced/Stopping Problems/parts/0 probability.tex b/Advanced/Stopping Problems/parts/0 probability.tex index e75b9da..b376b4c 100644 --- a/Advanced/Stopping Problems/parts/0 probability.tex +++ b/Advanced/Stopping Problems/parts/0 probability.tex @@ -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 $\Omega = \{ - \texttt{TTT},~ \texttt{TTH},~ \texttt{THT},~ - \texttt{THH},~ \texttt{HTT},~ \texttt{HTH},~ - \texttt{HHT},~ \texttt{HHH} + \texttt{TTT},~ \texttt{TTH},~ \texttt{THT},~ % spell:disable-line + \texttt{THH},~ \texttt{HTT},~ \texttt{HTH},~ % spell:disable-line + \texttt{HHT},~ \texttt{HHH} % spell:disable-line \}$, We can define a random variable $\mathcal{H}$ as \say{the number of heads in a throw of three coins}. \par As a function, $\mathcal{H}$ maps values in $\Omega$ to values in $\mathbb{Z}^+_0$ and is defined as: \begin{itemize} - \item $\mathcal{H}(\texttt{TTT}) = 0$ - \item $\mathcal{H}(\texttt{TTH}) = 1$ - \item $\mathcal{H}(\texttt{THT}) = 1$ - \item $\mathcal{H}(\texttt{THH}) = 2$ + \item $\mathcal{H}(\texttt{TTT}) = 0$ % spell:disable-line + \item $\mathcal{H}(\texttt{TTH}) = 1$ % spell:disable-line + \item $\mathcal{H}(\texttt{THT}) = 1$ % spell:disable-line + \item $\mathcal{H}(\texttt{THH}) = 2$ % spell:disable-line \item ...and so on. \end{itemize} @@ -70,7 +70,7 @@ the set of outcomes that produce that value. \par \vspace{2mm} 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{} diff --git a/Advanced/Stopping Problems/parts/2 secretary.tex b/Advanced/Stopping Problems/parts/2 secretary.tex index 2cd7c89..b0b26e7 100644 --- a/Advanced/Stopping Problems/parts/2 secretary.tex +++ b/Advanced/Stopping Problems/parts/2 secretary.tex @@ -191,7 +191,7 @@ what is the probability that we select the best candidate? \par Call this probability $\phi_n(k)$. \begin{solution} - Using \ref{seca} and \ref{secb}, this is straightfoward: + Using \ref{seca} and \ref{secb}, this is straightforward: \[ \phi_n(k) = \sum_{x = k}^{n}\left( \frac{1}{n} \times \frac{k-1}{x-1} \right) diff --git a/Intermediate/Instant Insanity/main.tex b/Intermediate/Instant Insanity/main.tex index 79236bc..6b292f0 100755 --- a/Intermediate/Instant Insanity/main.tex +++ b/Intermediate/Instant Insanity/main.tex @@ -120,7 +120,7 @@ \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 = 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) -- (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 + 4.5, 0 + 1.5) {\textbf{Red}}; \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 + 7.5, 0 + 4.5) {\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}}; % spell:disable-line \node[text = white] at (21 + 7.5, 0 - 1.5) {\textbf{Blue}}; \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 = black] at (0 + 4.5, -15 + 1.5) {\textbf{Grn}}; \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 + 7.5, -15 + 4.5) {\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}}; % spell:disable-line \node[text = white] at (0 + 7.5, -15 - 1.5) {\textbf{Blue}}; \filldraw [red] (21,-15) -- (24,-15) -- @@ -181,7 +181,7 @@ \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 = 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}}; \end{tikzpicture} @@ -270,7 +270,7 @@ \node[text = white] at (0.5, 5.5) {\textbf{Blue}}; \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}}; \end{tikzpicture} @@ -321,7 +321,7 @@ \node[text = white] at (0.5, 5.5) {\textbf{Blue}}; \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}}; \end{tikzpicture} \end{small} \end{center} @@ -400,7 +400,7 @@ \node[text = white] at (0.5, 5.5) {\textbf{Blue}}; \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}}; \end{tikzpicture} \end{small} \end{center} @@ -508,7 +508,7 @@ \node[text = white] at (0.5, 5.5) {\textbf{Blue}}; \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}}; \end{tikzpicture} \end{small} \end{center} @@ -611,7 +611,7 @@ \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, -1) {\textbf{Wht}}; + \node[text = black] at (4.5, -1) {\textbf{Wht}}; % spell:disable-line \end{tikzpicture} \end{center} @@ -657,10 +657,10 @@ \node[text = black] at (2.5, 0.5) {\textbf{Red}}; \node[text = white] at (1.5, -1) {\textbf{Red}}; \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 = 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{center} \bigskip diff --git a/Intermediate/Vectors 1/main.tex b/Intermediate/Vectors 1/main.tex index 745dcf8..cbbceb1 100755 --- a/Intermediate/Vectors 1/main.tex +++ b/Intermediate/Vectors 1/main.tex @@ -476,7 +476,10 @@ \section{Bonus} \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. + % spell:on \vfill diff --git a/Misc/Warm-Ups/furniture.tex b/Misc/Warm-Ups/furniture.tex index 534d76e..469c0a3 100755 --- a/Misc/Warm-Ups/furniture.tex +++ b/Misc/Warm-Ups/furniture.tex @@ -39,7 +39,7 @@ \item Rosewood \end{itemize} - The following contitions govern Irene's purchases: + The following conditions govern Irene's purchases: \begin{itemize} \item Any vanity she buys is Maple. \item Any rosewood item she buys is a sideboard. @@ -135,7 +135,7 @@ \problem{} Suppose the condition that Irene does not buy an oak table is 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? \begin{itemize} \item Irene buys an oak footstool. diff --git a/Misc/Warm-Ups/mario.tex b/Misc/Warm-Ups/mario.tex index e640531..fde7982 100755 --- a/Misc/Warm-Ups/mario.tex +++ b/Misc/Warm-Ups/mario.tex @@ -40,7 +40,11 @@ A 11-way tie is possible, with a top score of 28: \begin{itemize} \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 $10^\text{th}$; + % spell:on + \item Two players finish fifth twice and seventh twice, \item One player finishes sixth in each race. \end{itemize} diff --git a/Misc/Warm-Ups/regex.tex b/Misc/Warm-Ups/regex.tex index 3956d8a..b9e4e74 100644 --- a/Misc/Warm-Ups/regex.tex +++ b/Misc/Warm-Ups/regex.tex @@ -71,7 +71,9 @@ They specify exactly how many tokens to match: \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 + % spell:off \htexttt{ab\{2,\}a} will match any \texttt{ab...ba} with at least two \texttt{b}s. + % spell:on \vspace{5mm} diff --git a/resources/ormc_handout.cls b/resources/ormc_handout.cls index 83caa52..f7066b8 100755 --- a/resources/ormc_handout.cls +++ b/resources/ormc_handout.cls @@ -148,9 +148,9 @@ % Enum labels \renewcommand\theenumi{\@arabic\c@enumi} -\renewcommand\theenumii{\@alph\c@enumii} +\renewcommand\theenumii{\@alph\c@enumii} % spell:disable-line \renewcommand\theenumiii{\@roman\c@enumiii} -\renewcommand\theenumiv{\@Alph\c@enumiv} +\renewcommand\theenumiv{\@Alph\c@enumiv} % spell:disable-line \newcommand\labelenumi{\theenumi.} \newcommand\labelenumii{(\theenumii)} \newcommand\labelenumiii{\theenumiii.} diff --git a/typos.toml b/typos.toml index 05bc10f..1688011 100644 --- a/typos.toml +++ b/typos.toml @@ -1,5 +1,6 @@ [default] extend-words."LSAT" = "LSAT" +extend-words."ket" = "ket" extend-ignore-re = [ # spell:disable-line