Snakes!

lib/typst/local/handout/0.1.0/lib.typ

@@ -5,7 +5,7 @@
 #import "misc.typ": *
 #import "object.typ": definition, example, generic, problem, remark, theorem
 #import "solution.typ": (
-  if_no_solutions, if_solutions, if_solutions_else, instructornote,
+  if_no_solutions, if_solutions, if_solutions_else, instructornote, review_box,
   sample_solution, solution,
 )
 
@@ -30,12 +30,17 @@
   by: none,
   subtitle: none,
   short_warning: false,
+  page_numbers: true,
 ) = {
   set page(
     margin: 20mm,
     width: 8.5in,
     height: 11in,
-    footer: align(center, context counter(page).display()),
+    footer: {
+      if page_numbers {
+        align(center, context counter(page).display())
+      }
+    },
     footer-descent: 5mm,
   )
 
@@ -51,6 +56,8 @@
     justify: true,
   )
 
+  set math.mat(delim: "[")
+
   //
   // List style
   set list(

lib/typst/local/handout/0.1.0/solution.typ

@@ -106,6 +106,29 @@
   ))
 }
 
+#let review_box(title, content) = {
+  align(center, stack(
+    block(
+      width: 100%,
+      breakable: false,
+      fill: oblue,
+      stroke: oblue + 2pt,
+      inset: 1.5mm,
+      align(left, text(fill: white, weight: "bold", title)),
+    ),
+
+    block(
+      width: 100%,
+      height: auto,
+      breakable: false,
+      fill: oblue.lighten(80%).desaturate(10%),
+      stroke: oblue + 2pt,
+      inset: 3mm,
+      align(left, content),
+    ),
+  ))
+}
+
 
 #let instructornote(content) = {
   if_solutions(align(center, stack(

src/Advanced/Introduction to Quantum/src/parts/01 bits.tex

@@ -27,7 +27,7 @@ We'll represent this probabilistic bit's \textit{state} as a vector:
 $\left[\begin{smallmatrix}
 	p_0 \\ p_1
 \end{smallmatrix}\right]$ \par
-We do \textbf{not} assume this coin is fair, and thus $p_0$ might not equal $p_1$.
+We do \textbf{not} assume this coin is fair---$p_0$ might not equal $p_1$.
 
 \note{
 	This may seem a bit redundant: since $p_0 + p_1$, we can always calculate one probability given the other. \\
@@ -45,15 +45,16 @@ The simplest probabilistic bit states are of course $[0]$ and $[1]$, defined as
 	\item $[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
 	\item $[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$
 \end{itemize}
-That is, $[0]$ represents a bit that we known to be \texttt{0}, \par
+That is, $[0]$ represents a bit that we know to be \texttt{0}, \par
 and $[1]$ represents a bit we know to be \texttt{1}.
 
 \vfill
+\pagebreak
 
 
 \definition{}
-$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states: \par
-Every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$:
+$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states. This means that \par
+every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$:
 
 \begin{equation*}
 	\begin{bmatrix} p_0 \\ p_1 \end{bmatrix}
@@ -66,7 +67,6 @@ Every other probabilistic bit can be written as a \textit{linear combination} of
 
 
 \vfill
-\pagebreak
 
 \problem{}
 Every possible state of a probabilistic bit is a two-dimensional vector. \par
@@ -118,6 +118,10 @@ Draw all possible states on the axis below.
 
 
 
+%
+% MARK: measure
+%
+
 
 
 \section{Measuring Probabilistic Bits}
@@ -125,7 +129,7 @@ Draw all possible states on the axis below.
 
 \definition{}
 As we noted before, a probabilistic bit represents a coin we've tossed but haven't looked at. \par
-We do not know whether the bit is \texttt{0} or \texttt{1}, but we do know the probability of both of these outcomes. \par
+We do not know whether the bit is \texttt{0} or \texttt{1}, but we do know the probability of each outcome. \par
 
 \vspace{2mm}
 
@@ -135,7 +139,7 @@ knowledge of its state is updated to either $[0]$ or $[1]$, since we now certain
 \vspace{2mm}
 
 Since measurement changes what we know about a probabilistic bit, it changes the probabilistic bit's state.
-When we measure a bit, it's state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the
+When we measure a bit, its state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the
 bit vanishes. We \textit{cannot} recover the state $[x_0, x_1]$ from a measured probabilistic bit.
 
 
@@ -165,7 +169,10 @@ Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nic
 	\item If we measure $y$ first and observe \texttt{1}, \par
 	what is the probability of getting each of \texttt{00}, \texttt{01}, \texttt{10}, and \texttt{11}?
 \end{itemize}
-\note[Note]{$[x]$ and $[y]$ are column vectors, but I've written them horizontally to save space.}
+\note[Note]{
+	$[x]$ and $[y]$ are column vectors in the previous pages, and are still column vectors here. \par
+	I've written them horizontally to save space.
+}
 
 
 \vfill
@@ -189,7 +196,9 @@ What is the probability that $x$ and $y$ produce different outcomes?
 
 
 
-
+%
+% MARK: tensor product
+%
 
 \section{Tensor Products}
 
@@ -315,7 +324,7 @@ $?
 
 \problem{}
 What is the \textit{span} of the vectors we found in \ref{basistp}? \par
-In other words, what is the set of vectors that can be written as linear combinations of the vectors above?
+In other words, what is the set of all vectors that can be written as linear combinations of the vectors above?
 
 \vfill
 
@@ -387,7 +396,7 @@ Compute the following. Is the result what we'd expect?
 
 
 \problem{}<fivequant>
-Of course, writing $[0] \otimes [1]$ is a bit excessive. We'll shorten this notation to $[01]$. \par
+Writing $[0] \otimes [1]$ is tedious. We'll shorten this notation to $[01]$. \par
 
 \vspace{2mm}
 
@@ -429,9 +438,9 @@ Write the three-bit states $[0]$ through $[7]$ as column vectors. \par
 
 
 
-
-
-
+%
+% MARK: ops
+%
 
 
 
@@ -443,25 +452,6 @@ Write the three-bit states $[0]$ through $[7]$ as column vectors. \par
 Now that we can write probabilistic bits as vectors, we can represent operations on these bits
 with linear transformations---in other words, as matrices.
 
-\definition{}
-Consider the NOT gate, which operates as follows: \par
-\begin{itemize}
-	\item $\text{NOT}[0] = [1]$
-	\item $\text{NOT}[1] = [0]$
-\end{itemize}
-What should NOT do to a probabilistic bit $[x_0, x_1]$? \par
-If we return to our coin analogy, we can think of the NOT operation as
-flipping a coin we have already tossed, without looking at its state.
-Thus,
-\begin{equation*}
-	\text{NOT} \begin{bmatrix}
-		x_0 \\ x_1
-	\end{bmatrix} = \begin{bmatrix}
-		x_1 \\ x_0
-	\end{bmatrix}
-\end{equation*}
-
-
 \begin{hobox}{Review: Multiplying Vectors by Matrices}{black!10!white}{black!65!white}
 	\begin{equation*}
 		Av =
@@ -482,6 +472,10 @@ Thus,
 	Note that each element of $Av$ is the dot product of a row in $A$ and a column in $v$.
 \end{hobox}
 
+\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} interchangeably.
+
 \problem{}
 Compute the following product:
 \begin{equation*}
@@ -496,12 +490,23 @@ Compute the following product:
 
 \vfill
 
-\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} interchangeably.
-
-\pagebreak
-
+\definition{}
+Consider the NOT gate, which operates as follows: \par
+\begin{itemize}
+	\item $\text{NOT}[0] = [1]$
+	\item $\text{NOT}[1] = [0]$
+\end{itemize}
+What should NOT do to a probabilistic bit $[x_0, x_1]$? \par
+If we return to our coin analogy, we can think of the NOT operation as
+flipping a coin we have already tossed, without looking at its state.
+Thus,
+\begin{equation*}
+	\text{NOT} \begin{bmatrix}
+		x_0 \\ x_1
+	\end{bmatrix} = \begin{bmatrix}
+		x_1 \\ x_0
+	\end{bmatrix}
+\end{equation*}
 
 \problem{}
 Find the matrix that represents the NOT operation on one probabilistic bit.
@@ -516,6 +521,8 @@ Find the matrix that represents the NOT operation on one probabilistic bit.
 
 \vfill
 
+\pagebreak
+
 
 \problem{Extension by linearity}
 Say we have an arbitrary operation $M$. \par

src/Advanced/Introduction to Quantum/src/parts/02 qubit.tex

@@ -4,7 +4,8 @@ Quantum bits (or \textit{qubits}) are very similar to probabilistic bits, but ha
 probabilities are replaced with \textit{amplitudes}.
 
 \vspace{2mm}
-Of course, a qubit can take the values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par
+
+As before a qubit can take the extremal values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par
 Like probabilistic bits, a quantum bit is written as a linear combination of $\ket{0}$ and $\ket{1}$:
 \begin{equation*}
 	\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}
@@ -19,8 +20,8 @@ $\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one
 
 \vspace{2mm}
 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 $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
+When we work with qubits, we will write $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ as $\ket{0}$
+and $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ as $\ket{1}$.
 
 
 \vspace{8mm}
@@ -28,7 +29,7 @@ and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
 Recall that probabilistic bits are subject to the restriction that $p_0 + p_1 = 1$. \par
 Quantum bits have a similar condition: $\psi_0^2 + \psi_1^2 = 1$. \par
 Note that this implies that $\psi_0$ and $\psi_1$ are both in $[-1, 1]$. \par
-Quantum amplitudes may be negative, but probabilistic bit probabilities cannot.
+Quantum amplitudes may be negative!
 
 \vspace{2mm}
 
@@ -89,8 +90,8 @@ Write $\ket{\psi}$ as a linear combination of $\ket{0}$ and $\ket{1}$.
 
 
 \definition{Measurement I}
-Just like a probabilistic bit, we must observed $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par
-If we were to measure $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$, we'd observe either $\ket{0}$ or $\ket{1}$, \par
+Just like a probabilistic bit, we must observe $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par
+If we measure $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$, we will see either $\ket{0}$ or $\ket{1}$,
 with the following probabilities:
 \begin{itemize}[itemsep = 2mm, topsep = 2mm]
 	\item $\mathcal{P}(\ket{1}) = \psi_1^2$
@@ -117,7 +118,7 @@ leaving no trace of the previous superposition. \par
 
 \problem{}
 \begin{itemize}
-	\item What is the probability we observe $\ket{0}$ when we measure $\ket{\psi}$? \par
+	\item What is the probability that we observe $\ket{0}$ when we measure $\ket{\psi}$? \par
 	\item What can we observe if we measure $\ket{\psi}$ a second time? \par
 	\item What are these probabilities for $\ket{\varphi}$?
 \end{itemize}
@@ -235,6 +236,10 @@ Consider the following qubit states:
 \pagebreak
 
 
+%
+% MARK: ops
+%
+
 \section{Operations on One Qubit}
 
 We may apply transformations to qubits just as we apply transformations to probabilistic bits.
@@ -277,7 +282,7 @@ Find the matrix $X$.
 \vfill
 
 \problem{}
-What is $X\ket{+}$ and $X\ket{-}$? \par
+What are $X\ket{+}$ and $X\ket{-}$? \par
 \hint{Remember that all matrices are linear maps. What does this mean?}
 
 \begin{solution}
@@ -331,11 +336,21 @@ As we noted earlier, any rotation about the center is a valid quantum gate. \par
 Let's derive all transformations of this form.
 \begin{itemize}[itemsep = 1mm]
 	\item Let $U_\phi$ be the matrix that represents a counterclockwise rotation of $\phi$ degrees. \par
-	What is $U\ket{0}$ and $U\ket{1}$?
+	What are $U\ket{0}$ and $U\ket{1}$?
 
 	\item Find the matrix $U_\phi$ for an arbitrary $\phi$.
 \end{itemize}
 
+
+\begin{solution}
+	\begin{equation*}
+		\begin{bmatrix}
+			cos(\theta) & -sin(\theta) \\
+			sin(\theta) & cos(\theta)
+		\end{bmatrix}
+	\end{equation*}
+\end{solution}
+
 \vfill
 
 
@@ -343,5 +358,10 @@ Let's derive all transformations of this form.
 Say we have a qubit that is either $\ket{+}$ or $\ket{-}$. We do not know which of the two states it is in. \par
 Using one operation and one measurement, how can we find out, for certain, which qubit we received? \par
 
+
+\begin{solution}
+	Use a 45 degree ccw rotation.
+\end{solution}
+
 \vfill
 \pagebreak

src/Advanced/Introduction to Quantum/src/parts/03 two qubits.tex

@@ -60,7 +60,7 @@ $\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3
 Again, consider the two-qubit state
 $\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$ \par
 If we measure the first qubit of $\ket{\psi}$ and get $\ket{0}$, what is the resulting state of $\ket{\psi}$? \par
-What would the state be if we'd measured $\ket{1}$ instead?
+What would that state be if we'd measured $\ket{1}$ instead?
 
 \vfill
 
@@ -85,7 +85,7 @@ Say we measure the first two qubits and get $\ket{00}$. What is the resulting st
 
 
 \definition{Entanglement}
-Some product states can be factored into a tensor product of individual qubit states. For example,
+Some compound states can be factored into a tensor product of individual qubit states. For example,
 \begin{equation*}
 	\frac{1}{2} \bigl(\ket{00} + \ket{01} + \ket{10} + \ket{11}\bigr)
 	= \frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr) \otimes

src/Advanced/Introduction to Quantum/src/parts/04 logic gates.tex

@@ -1,40 +1,42 @@
 \section{Logic Gates}
 
-\definition{Matrices}
-Throughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix,
-and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear
-map, we can write it as follows:
-\begin{equation*}
-		f\left(
-			\ket{x}
-		\right)
-	=
-		\begin{bmatrix}
-			m_1 & m_2 \\
-			m_3 & m_4
-		\end{bmatrix}
-		\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
-	=
-		\left[
-		\begin{matrix}
-			m_1x_1 + m_2x_2 \\
-			m_3x_1 + m_4x_2
-		\end{matrix}
-		\right]
-\end{equation*}
-
+%\definition{Matrices}
+%cThroughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix,
+%and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear
+%map, we can write it as follows:
+%\begin{equation*}
+%		f\left(
+%			\ket{x}
+%		\right)
+%	=
+%		\begin{bmatrix}
+%			m_1 & m_2 \\
+%			m_3 & m_4
+%		\end{bmatrix}
+%		\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
+%	=
+%		\left[
+%		\begin{matrix}
+%			m_1x_1 + m_2x_2 \\
+%			m_3x_1 + m_4x_2
+%		\end{matrix}
+%		\right]
+%\end{equation*}
+%
+%A classical logic gate is a linear map from $\{0,1\}^m$ to $\{0,1\}^n$
 
 \definition{}
-Before we discussing multi-qubit quantum gates, we need to review to classical logic. \par
-Of course, a classical logic gate is a linear map from $\{0,1\}^m$ to $\{0,1\}^n$
+Before we discussing multi-qubit quantum gates, we need to review classical logic. \par
+In this section, let's return to probabilistic bits.
+
 
 
 \problem{}<notgatex>
 The \texttt{not} gate is a map defined by the following table: \par
 
 \begin{itemize}
-	\item $X\ket{0} = \ket{1}$
-	\item $X\ket{1} = \ket{0}$
+	\item $X[0] = [1]$
+	\item $X[1] = [0]$
 \end{itemize}
 
 Write the \texttt{not} gate as a matrix that operates on single-bit vector states. \par
@@ -93,11 +95,11 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
 
 		\vspace{2mm}
 		For example, if we look at the first column of $A$ (which is $[1, 0]$), we see: \par
-		$A\ket{00} = A[1,0,0,0] = [1,0] = \ket{0}$
+		$A[00] = A[1,0,0,0] = [1,0] = [0]$
 
 		\vspace{2mm}
 		Also with the last column (which is $[0,1]$): \par
-		$A\ket{00} = A[0,0,0,1] = [0,1] = \ket{1}$
+		$A[00] = A[0,0,0,1] = [0,1] = [1]$
 	\end{instructornote}
 \end{solution}
 
@@ -244,10 +246,10 @@ Say we want to invert the first bit of a two-bit state. That is, we want a trans
 
 In other words, we want a matrix $T$ satisfying the following equalities:
 \begin{itemize}
-	\item $T\ket{00} = \ket{10}$
-	\item $T\ket{01} = \ket{11}$
-	\item $T\ket{10} = \ket{00}$
-	\item $T\ket{11} = \ket{01}$
+	\item $T[00] = [10]$
+	\item $T[01] = [11]$
+	\item $T[10] = [00]$
+	\item $T[11] = [01]$
 \end{itemize}
 
 
@@ -256,8 +258,8 @@ In other words, we want a matrix $T$ satisfying the following equalities:
 Find the matrix that corresponds to the above transformation. \par
 \hint{
 	Remember that
-	$\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and
-	$\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\
+	$[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and
+	$[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\
 	Also, we found earlier that $X = \left[\begin{smallmatrix} 0 && 1 \\ 1 && 0 \end{smallmatrix}\right]$,
 	and of course $I = \left[\begin{smallmatrix} 1 && 0 \\ 0 && 1 \end{smallmatrix}\right]$.
 }
@@ -279,25 +281,26 @@ Find the matrix that corresponds to the above transformation. \par
 We could draw the above transformation as a combination $X$ and $I$ (identity) gate:
 \begin{center}
 \begin{tikzpicture}[scale=0.8]
-	\node[qubit] (a) at (0, 0) {$\ket{0}$};
-	\node[qubit] (b) at (0, -1) {$\ket{0}$};
+	\node[qubit] (a) at (0, 0) {$[0]$};
+	\node[qubit] (b) at (0, -1) {$[0]$};
 
-	\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
-	\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
+	\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$[1]$};
+	\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$[0]$};
 
 	\qubox{a}{1}{a}{2}{$X$}
 	\qubox{b}{1}{b}{2}{$I$}
 \end{tikzpicture}
 \end{center}
 
-We can even omit the $I$ gate, since we now know that transformations affect the whole state: \par
+We can even omit the $I$ gate. We know that transformations affect the whole state,
+and can assume the empty space uncer $X$ implies $I$. \par
 \begin{center}
 \begin{tikzpicture}[scale=0.8]
-	\node[qubit] (a) at (0, 0) {$\ket{0}$};
-	\node[qubit] (b) at (0, -1) {$\ket{0}$};
+	\node[qubit] (a) at (0, 0) {$[0]$};
+	\node[qubit] (b) at (0, -1) {$[0]$};
 
-	\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
-	\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
+	\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$[1]$};
+	\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$[0]$};
 
 	\qubox{a}{1}{a}{2}{$X$}
 \end{tikzpicture}

src/Advanced/Introduction to Quantum/src/parts/06 hxh.tex

@@ -1,7 +1,7 @@
 \section{HXH}
 
 Let's return to the quantum circuit diagrams we discussed a few pages ago. \par
-Keep in mind that we're working with quantum gates and proper half-qubits---not classical bits, as we were before.
+Keep in mind that we're working with quantum gates and proper qubits---not classical bits, as we were before.
 
 \definition{Controlled Inputs}
 A \textit{control input} or \textit{inverted control input} may be attached to any gate. \par
@@ -289,7 +289,7 @@ $\ket{-} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} - \ket{1}\Bigr)$
 \pagebreak
 
 \generic{Remark:}
-Now, consider the following circuit:
+Consider the following circuit:
 
 \begin{center}
 	\begin{tikzpicture}[scale=0.8]
@@ -386,21 +386,21 @@ but the state $\ket{cd}$ on the right is $\ket{11} = [0,0,0,1]$. \par
 \vspace{4mm}
 
 How does this make sense? \par
-Remember that a two-bit quantum state is \textit{not} equivalent to a pair of one-qubit quantum states.
-We must treat a multi-qubit state as a single unit.
+Remember that a two-bit quantum state is \textit{not} equivalent to a pair of disjoint one-qubit quantum states.
+We cannot treat a multi-qubit state as a combination of $n$ independent bits.
 
-Recall that a two-bit state $\ket{ab}$ comes with four probabilities:
-$\mathcal{P}(\texttt{00})$, $\mathcal{P}(\texttt{01})$, $\mathcal{P}(\texttt{10})$, and $\mathcal{P}(\texttt{11})$.
+\vspace{2mm}
+
+A two-bit state $\ket{ab}$ comes with four probabilities:
+$\mathcal{P}(\texttt{00})$, $\mathcal{P}(\texttt{01})$, $\mathcal{P}(\texttt{10})$, and $\mathcal{P}(\texttt{11})$. \par
 If we change the probabilities of only $\ket{a}$, \textit{all four of these change!}
 
 
 \vfill
 
-Because of this fact, \say{controlled gates} may not work as you expect. They may seem
-to \say{read} their controlling qubit without affecting its state, but remember---a
-controlled gate still affects the \textit{entire} state. As we noted before, it is
-not possible to apply a transformation to one bit of a quantum state.
-
+Because of this fact, \say{controlled gates} behave in a somewhat counterintuitive way.
+They do not simply \say{read} their controlling qubit without affecting its state---
+they mutate the entire state they are applied to, and may change all its bits!
 
 \begin{center}
 	\begin{tikzpicture}[scale=1]

src/Advanced/Introduction to Quantum/src/parts/07 superdense.tex

@@ -12,7 +12,7 @@ Consider the following entangled two-qubit states, called the \textit{bell state
 The probabilistic bits we get when measuring any of the above may be called \textit{anticorrelated bits}. \par
 If we measure the first bit of any of these states and observe $1$, what is the resulting compound state? \par
 What if we observe $0$ instead? \par
-Do you see why we can call these bits anticorrelated?
+Do you see why we call these bits \say{anticorrelated}?
 
 \vfill
 

src/Advanced/Symmetric Groups/main.typ

@@ -20,7 +20,7 @@
 = Bonus problems
 
 #problem()
-Show that $x in ZZ^+$ has a multiplicative inverse mod $n$ iff $gcd(x, n) = 1$
+Show that $x in ZZ^+$ has a multiplicative inverse mod $n$ if and only if $gcd(x, n) = 1$
 
 #v(1fr)
 

src/Advanced/Symmetric Groups/parts/00 intro.typ

@@ -45,7 +45,7 @@ How many permutations of $n$ objects are there?
 #v(1fr)
 
 #problem()
-What map corresponds to the permutation that produces the array `312` from the array `123`?
+What map corresponds to the permutation that produces the array `231` from the array `123`?
 
 #v(1fr)
 

src/Advanced/Symmetric Groups/parts/01 cycle.typ

@@ -100,7 +100,7 @@ How about $[321][213][231]$? \
 Rewrite these compositions as one permutation in square brackets.
 
 #solution([
-  - $[1324][4321]$ is $[4321]$
+  - $[1324][4321]$ is $[4231]$
   - $[321][213][231]$ is $[123]$
 ])
 
@@ -501,7 +501,7 @@ List all other ways to write this cycle. \
 
 
 #definition("Inverse")
-The _inverse_ of a permitation $f$ is a permutation $g$ that "un-does" $f$. \
+The _inverse_ of a permutation $f$ is a permutation $g$ that "un-does" $f$. \
 This means that $g(f(x)) = x$ for all $x$.
 
 #problem()
@@ -550,7 +550,7 @@ Show that any cycle $(123...n)$ is equal to the product $(12)(23)...(n-1, n)$.
 Write $(7126453)$ as a product of transpositions. \
 
 #solution[
-  Move elements one at a time, and using the last position as temporary storage.
+  Move elements one at a time, using the last position as temporary storage.
 
   We get $(71)(72)(76)(74)(75)(73)$.
   Other solutions are possible. \
@@ -589,7 +589,7 @@ Show that any permutation is a product of transpositions of the form $(1, k)$. \
 
 
 #problem(label: "oneplustrans")
-Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$.
+Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$ whenever $a + 1 != b$.
 
 #solution[
   This is the same as @onetrans,

src/Advanced/Symmetric Groups/parts/02 groups.typ

@@ -2,20 +2,16 @@
 #import "@preview/cetz:0.4.2"
 #import "../macros.typ": *
 
-= Groups (review)
+= Groups
 
 #definition()
 Before we continue, we must introduce a bit of notation:
 - $S_n$ is the set of permutations on $n$ objects.
 - $ZZ_n$ is the set of integers mod $n$.
 
-- $ZZ_n^times$ is the set of integers mod $n$ with multiplicative inverses. \
-  In other words, it is the set of integers smaller than $n$ and coprime to $n$.#footnote[We proved this in another handout, but you may take it as fact here.] \
-  For example, $ZZ_12^times = {1, 5, 7, 11}$.
-
 #problem()
 What are the elements of $S_3$? #hint[Use cycle notation] \
-How about $ZZ_17^times$?
+How about $ZZ_8$?
 
 #v(1fr)
 
@@ -47,6 +43,16 @@ What is the group with the fewest number of elements?
   Verifying that the trivial group is a group is trivial.
 ]
 
+
+#definition()
+$ZZ_n^times$ is the set of integers mod $n$ with multiplicative inverses. \
+We can prove that this is the set of integers smaller than $n$ and coprime to $n$. \
+For example, $ZZ_12^times = {1, 5, 7, 11}$.
+
+#problem()
+What are the elements of $ZZ^times_8$? \
+How about $ZZ^times_23$? #hint[23 is prime.]
+
 #v(1fr)
 #pagebreak()
 
@@ -66,8 +72,11 @@ Show that $S_n$ is a group under composition.
 
 #problem()
 Let $(G, *)$ be a group with finitely many elements, and let $a in G$. \
-Show that there is an $n$ in $in ZZ^+$ so that $a^n = e$ \
-#hint[$a^n = a * a * ... * a$ repeated $n$ times.]
+Show that there is an $n$ in $in ZZ$ so that $a^n = e$ \
+#hint[
+  $a^n = a * a * ... * a$ repeated $n$ times. \
+  $a^(-n) = a^(-1) * a^(-1) * ... * a^(-1)$, where $a^(-1)$ is the inverse of $a$. \
+]
 
 #v(2mm)
 
@@ -111,7 +120,7 @@ Then, find all generators of $(ZZ_5, +)$
 How many groups have only one generator?
 
 #solution[
-  Only one: the trivial group. The inverse of a generator is also a generator!
+  Two: the trivial group and $(ZZ_2, +)$.
 ]
 
 #v(1fr)

src/Warm-Ups/Snakes/0.svg

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src/Warm-Ups/Snakes/main.typ

@@ -0,0 +1,89 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.4.2"
+
+#show: handout.with(
+  title: [Warm-Up: Snakes!],
+  by: "Mark",
+  page_numbers: false,
+)
+
+#box(place(box(width: 100%, height: 8in, align(horizon, image(
+  "7.svg",
+  width: 360mm,
+)))))
+
+#problem()
+Evaluate this integral. \
+All integrals are of the form $integral_a^b 1 #h(1mm) d x$.
+
+#if_solutions(pagebreak())
+
+#solution[
+  This integral is drawn recursively with the following code: \
+
+  #v(2mm)
+
+  #let snake(depth, x, y) = {
+    if depth == 0 {
+      math.attach(math.integral, br: x, tr: y)
+    } else {
+      let bot = snake(depth - 1, x, "1")
+      let top = snake(depth - 1, "0", y)
+      snake(depth - 1, bot, top)
+    }
+  }
+
+  ```typst
+  #let snake(depth, x, y) = {
+    if depth == 0 {
+      $integral_#x ^#y$
+    } else {
+      let bot = snake(depth - 1, x, "1")
+      let top = snake(depth - 1, "0", y)
+      snake(depth - 1, bot, top)
+    }
+  }
+  ```
+
+  #v(5mm)
+
+  For example:
+
+  $ snake\(0, x, y) = #snake(0, "x", "y") $
+
+  #v(2mm)
+
+  $ snake\(1, x, y) = #snake(1, "x", "y") $
+
+  #v(2mm)
+
+  $ snake\(2, x, y) = #snake(2, "x", "y") $
+
+  In other words, we want $f_7 (0, 1)$, where...
+  - $f_0(x, y) := integral_x^y$
+  - $f_(n+1)(x, y) := f_n [f_n (x, 1), f_n (0, y)]$
+
+  #v(5mm)
+  Expanding a few iterations, we find:
+  - $f_0(x, y) = integral_x^y = y-x$
+  - $f_1(x, y) = f_0 [f_0 (x, 1), f_0 (0, y)] = y+x-1$
+  - $f_2(x, y) = f_1 [f_1 (x, 1), f_1 (0, y)] = y+x-2$
+
+  #v(5mm)
+
+  We can use induction to show that this pattern continues. \
+  If $g_n = y+x+c$, then $g_(n+1) = y+x+(3c+1)$.
+
+
+  #v(5mm)
+  Finally, use this recusion to find that
+  $f_0, f_1, ..., f_7 = 1, 0, -1, -4, -13, -40, -121, -364$
+
+  One can also find an explicit formula for $g_n$:
+
+  $
+    f_(n+1) = g_n & = x + y + 3^n c + 3^0 + 3^1 + ... + 3^n \
+                  & = x + y + 3^n c + sum_(i=0)^n 3^i       \
+                  & = x + y + 3^n c + (3^(n+1) + 1)/2
+  $
+]

src/Warm-Ups/Snakes/meta.toml

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

src/Warm-Ups/Snakes/svg.typ

@@ -0,0 +1,57 @@
+#import "@preview/cetz:0.4.2"
+
+// Compile with:
+// typst compile --package-path "../../../lib/typst/" svg.typ --format svg
+// You'll need to recompile typst with a higher recursion limit.
+
+#set page(
+  fill: none,
+  width: auto,
+  height: auto,
+  margin: 0mm,
+);
+
+#let cetz_snake(depth, b, t) = {
+  import cetz.draw: *
+
+  if depth == 0 {
+    content((0, 0), math.inline(math.integral))
+
+    group({
+      translate(x: 0.15, y: -0.15)
+      b
+    })
+
+    group({
+      translate(x: 0.15, y: 0.15)
+      t
+    })
+  } else {
+    let bot = group({
+      translate(x: 0.15, y: -0.15)
+      cetz_snake(depth - 1, b, content((0, 0), text("1", size: 3mm)))
+    })
+
+    let top = group({
+      translate(x: 0.15, y: 0.15)
+      cetz_snake(depth - 1, content((0, 0), text("0", size: 3mm)), t)
+    })
+
+    cetz_snake(depth - 1, bot, top)
+  }
+}
+
+
+#{
+  import cetz.draw: *
+
+  // any bigger than 7 overflows the stack.
+  // any bigger than 5 exceeds typst's recursion limit.
+  // You'll have to increase those limits and recompile to
+  // build this document.
+  let depth = 7
+
+  let b = content((0, 0), text("0", size: 3mm))
+  let t = content((0, 0), text("1", size: 3mm))
+  cetz.canvas(cetz_snake(depth, b, t))
+};