More fixes

CodingAndCryptography/00_modelling_communication.tex

@@ -27,7 +27,7 @@ \section{Modelling communication}
 \end{definition}
 
 \begin{definition}[Discrete Memoryless Channel]
-    A \vocab{discrete memoryless channel} (DMC) is a channel where $p_{ij} = \prob{b_j \text{ received} \mid a_i \text{ sent}}$ are the same for each channel use, and independent of all past and future uses of the channel.
+    A \vocab{discrete memoryless channel} (DMC) is a channel where $p_{ij} = \mathbb{P}(b_j \text{ received} \mid a_i \text{ sent})$ are the same for each channel use, and independent of all past and future uses of the channel.
     Its \vocab{channel matrix} is the $r \times s$ stochastic matrix $P = (p_{ij})$.
 \end{definition}
 

CodingAndCryptography/01_noiseless_coding.tex

@@ -195,7 +195,7 @@ \subsection{Gibbs' inequality}
 
 \subsection{Optimal codes}
 Let $\mathcal A = \qty{\mu_1, \dots, \mu_m}$ be an alphabet of $m \geq 2$ messages, and let $\mathcal B$ be an alphabet of length $a \geq 2$.
-Let $X$ be a r.v. taking values in $A$ with probabilities $p_1, \dots, p_m$.
+Let $X$ be a r.v. taking values in $\mathcal{A}$ with probabilities $p_1, \dots, p_m$.
 
 \begin{definition}[Optimal Code]
     A code $c \colon \mathcal A \to \mathcal B^\star$ is called \vocab{optimal} if it has the smallest possible expected word length $\sum p_i \ell_i = \expect{S}$ among all decipherable codes.

QuantumInfoAndComputing/03_quantum_cryptography.tex

@@ -44,19 +44,20 @@ \subsection{The BB84 protocol}
     \ket{-} & x_i = 1; y_i = 1
 \end{cases} \]
 Alice sends the qubits $\ket{\psi_{xy}}$ to Bob with $m$ uses of the quantum channel.
-The qubits received are not necessarily in the state $\ket{\psi_{xy}}$ due to noise or malicious manipulation of the channel.
+The qubits received are not necessarily in the state $\ket{\psi_{xy}}$ due to noise or malicious manipulation of the channel, but for now assume they are.
+
 Bob then generates an $m$-bit string $y' = y_1' \dots y_m' \in \qty{0,1}^m$ uniformly at random.
 If $y_i' = 0$, he measures the $i$th qubit in the basis $\mathcal B_0 = \qty{\ket{0}, \ket{1}}$.
 If $y_i' = 1$, he acts on the $i$th qubit by the Hadamard gate and then measures in $\mathcal B_0$.
 Equivalently, he measures the $i$th qubit in the basis $\mathcal B_1 = \qty{\ket{+}, \ket{-}}$.
 Let the sequence of outcomes be $x' = x_1' \dots x_m' \in \qty{0,1}^m$.
 
-If $y_i' = y_i$, we have $x_i' = x_i$.
-Indeed, suppose $y_i' = 0 = y_i$.
-Then $\ket{\pi_{x_i y_i}} \in \mathcal B_0$, and Bob measures in basis $\mathcal B_0$, so he can determine $x_i$ with probability 1.
-If $y_i' = 1 = y_i$, $\ket{\pi_{x_i y_i}} \in \mathcal B_1$, and Bob measures in basis $\mathcal B_1$.
+If $y_i' = y_i$, we have $x_i' = x_i$ else $x_i$ and $x_i'$ are uncorrelated.
+% Indeed, suppose $y_i' = 0 = y_i$.
+% Then $\ket{\psi_{x_i y_i}} \in \mathcal B_0$, and Bob measures in basis $\mathcal B_0$, so he can determine $x_i$ with probability 1.
+% If $y_i' = 1 = y_i$, $\ket{\psi_{x_i y_i}} \in \mathcal B_1$, and Bob measures in basis $\mathcal B_1$.
 
-Now, Alice and Bob compare their values of $y$ and $y'$ over the classical channel, and discard all $x_i$ and $x_i'$ for which $y_i \neq y_i'$.
+Now, Alice and Bob publically compare their values of $y$ and $y'$ over the classical channel, and discard all $x_i$ and $x_i'$ for which $y_i \neq y_i'$.
 The remaining $x_i$ and $x_i'$ match, given that Bob receives $\ket{\psi_{xy}}$ exactly, and this forms the shared private key $\widetilde x = \widetilde x'$.
 The average length of $\widetilde x$ is $\frac{m}{2}$.
 

QuantumInfoAndComputing/04_quantum_computation.tex

@@ -217,7 +217,7 @@ \subsubsection{Application: Periodicity Determination}
     &= \frac{1}{\sqrt{A}} \frac{1}{\sqrt{N}} \sum_{j=0}^{A-1} \sum_{y=0}^{N-1} \omega^{(x_0 + jr)y} \ket{y} \\
     &= \frac{1}{\sqrt{NA}} \sum_{y=0}^{N-1} \omega^{x_0 y} \underbrace{\qty[\sum_{j=0}^{A-1} \qty(\omega^{ry})^j]}_S \ket{y}
 \end{align*}
-Note that
+Note that $Ar = N$ and
 \[ S = \begin{cases}
     A & \text{if } \omega^{ry} = 1 \\
     \frac{1 - \omega^{ryA}}{1 - \omega^{ry}} = 0 & \text{otherwise}