Finish off QIC

LogicAndSetTheory/05_set_theory.tex

@@ -628,7 +628,7 @@ \subsection{The universe of sets}
 \begin{remark}
     If $x \in V_\alpha$ then $x \subseteq V_\alpha$ by \cref{lem:5-6}. \\
     If $x \subseteq V_\alpha$ then $x \in \mathcal{P}(V_\alpha) = V_{\alpha+1}$.
-    If $\exists \alpha \in ON$ s.t. $x \subset V_\alpha$ define the \vocab{rank} of $x$ to be the least such $\alpha$.
+    If $\exists \alpha \in ON$ s.t. $x \subseteq V_\alpha$ define the \vocab{rank} of $x$ to be the least such $\alpha$.
 
     For example, the rank of $\emptyset$ is 0, the rank of 1 is 1, the rank of $\omega$ is $\omega$, and in general the rank of any ordinal $\alpha$ is $\alpha$.
     Intuitively, the rank of a set is the time at which it was created.

QuantumInfoAndComputing/04_quantum_computation.tex

@@ -378,7 +378,7 @@ \subsection{Grover's algorithm}
 Given a black box which computes $I_{x_0}$ for some $x_0 \in B_n$, we wish to determine $x_0$ with the least amount of queries.
 We will now describe Grover's algorithm.
 We begin with the equal superposition state $\ket{\psi_0} = \frac{1}{\sqrt{2^n}} \sum_{x \in B_n} \ket{x}$.
-Consider \vocab{Grover's iteration operator} $Q = -H_n I_0 H_n I_{x_0}$ where $H_n = H^{\otimes n}$.
+Consider \vocab{Grover's iteration operator} $Q = -H_n I_0 H_n I_{x_0} = -I_{\psi_0} I_{x_0}$\footnote{$H_n I_0 H_n = I_{H \ket{0}} = I_{\psi_0}$.} where $H_n = H^{\otimes n}$.
 Note that $Q$ is real-valued, so acts geometrically on the real-valued vector $\ket{\psi_0}$ in real Euclidean space.
 It has the following properties.
 \begin{enumerate}
@@ -459,17 +459,19 @@ \subsection{Grover's algorithm for multiple items}
 Hence, as before, $Q_G$ causes the desired rotation through $2\alpha$ in this plane.
 The probability of finding a single good item is $\abs{\ip{\psi}{\psi_G}}^2$, as $\ket{\psi} = a \ket{\psi_G} + b \ket{\psi_B}$.
 
-Suppose now that $r$ is unknown.
-In this case, we start with $\ket{\psi_0}$ and repeatedly apply $Q$ to rotate $\ket{\psi_0}$ to $\ket{\psi_G}$ as before.
-However, we do not know how many iterations of $Q$ to apply, since this depends on $r$.
+\begin{aside}{$r$ unknown - Non Examinable}
+    Suppose now that $r$ is unknown.
+    In this case, we start with $\ket{\psi_0}$ and repeatedly apply $Q$ to rotate $\ket{\psi_0}$ to $\ket{\psi_G}$ as before.
+    However, we do not know how many iterations of $Q$ to apply, since this depends on $r$.
 
-If $r \ll N$, we choose $K$ uniformly at random in $\qty(0, \frac{\pi}{4}\sqrt{N})$, and apply $K$ iterations of $Q$.
-We measure the final state $\ket{\psi^K}$ to obtain $x$, and check if $f(x) = 1$ or not.
-Note that each iteration causes a rotation of $2\alpha$ where $\sin \alpha = \frac{\sqrt{r}}{\sqrt{N}}$ so $2\alpha \approx 2\frac{\sqrt{r}}{\sqrt{N}}$.
-Choosing $K$ therefore implicitly chooses a random angle in the range $\qty(0, \frac{\pi}{2} \sqrt{r})$.
-Now, if the final rotated state $\ket{\psi}$ makes an angle within $\pm \frac{\pi}{4}$ with $\ket{\psi_0}$, the probability of locating a good item is $\abs{\ip{\psi}{\psi_0}}^2 \geq \cos^2 \frac{\pi}{4} = \frac{1}{2}$.
-Since for every quadrant in the plane $\mathcal P_G$, half of the angles are within $\pm \frac{\pi}{4}$ from the $y$-axis, the randomised procedure using $O(\sqrt{N})$ queries will locate a good item with probability approximately $\frac{1}{4}$.
-The procedure can then be repeated to reduce the error probability to an acceptable level.
+    If $r \ll N$, we choose $K$ uniformly at random in $\qty(0, \frac{\pi}{4}\sqrt{N})$, and apply $K$ iterations of $Q$.
+    We measure the final state $\ket{\psi^K}$ to obtain $x$, and check if $f(x) = 1$ or not.
+    Note that each iteration causes a rotation of $2\alpha$ where $\sin \alpha = \frac{\sqrt{r}}{\sqrt{N}}$ so $2\alpha \approx 2\frac{\sqrt{r}}{\sqrt{N}}$.
+    Choosing $K$ therefore implicitly chooses a random angle in the range $\qty(0, \frac{\pi}{2} \sqrt{r})$.
+    Now, if the final rotated state $\ket{\psi}$ makes an angle within $\pm \frac{\pi}{4}$ with $\ket{\psi_G}$, the probability of locating a good item is $\abs{\ip{\psi}{\psi_G}}^2 \geq \cos^2 \frac{\pi}{4} = \frac{1}{2}$.
+    Since for every quadrant in the plane $\mathcal P_G$, half of the angles are within $\pm \frac{\pi}{4}$ from the $y$-axis, the randomised procedure using $O(\sqrt{N})$ queries will locate a good item with probability approximately $\frac{1}{4}$.
+    The procedure can then be repeated to reduce the error probability to an acceptable level.
+\end{aside}
 
 \subsection{\texorpdfstring{$\mathsf{NP}$}{NP} problems}
 A \vocab{verifier} $V$ for a language $L$ is a computation with two inputs $w, c$ such that
@@ -512,16 +514,16 @@ \subsection{Shor's algorithm}
 If $b = 1$, then $a, N$ are coprime.
 \begin{theorem}[Euler's theorem]
     Let $a, N$ be coprime.
-    Then there exists $1 < r < N$ such that $a^r \equiv 1$ mod $N$.
+    Then there exists $1 < r < N$ s.t. $a^r \equiv 1$ mod $N$.
     A minimal such $r$ is called the \vocab{order} of $a$ modulo $N$.
 \end{theorem}
-Consider the \vocab{modular exponentiation function} $f \colon \mathbb Z \to \faktor{\mathbb Z}{n\mathbb Z}$ such that $f(k) = a^k$ mod $N$.
+Consider the \vocab{modular exponentiation function} $f \colon \mathbb Z \to \faktor{\mathbb Z}{N\mathbb Z}$ s.t. $f(k) = a^k$ mod $N$.
 This function satisfies $f(k_1 + k_2) = f(k_1)f(k_2)$.
-$f$ is periodic with period $r$, and is injective within each period.
+$f$ is periodic with period $r$, and is injective within each period as $r$ minimal.
 
 Suppose that we can find $r$, and suppose $r$ is even.
 Then $a^r - 1 \equiv (a^{\frac{r}{2}}+1)(a^{\frac{r}{2}}-1) \equiv 0$ mod $N$.
-Note that $N \nmid (a^{\frac{r}{2}}-1)$ since $r$ was minimal such that $a^r \equiv 1$ mod $N$.
+Note that $N \nmid (a^{\frac{r}{2}}-1)$ since $r$ was minimal s.t. $a^r \equiv 1$ mod $N$.
 If $N \nmid (a^{\frac{r}{2}}+1)$, then $N$ must have some prime factors in $(a^{\frac{r}{2}}+1)$ and some in $(a^{\frac{r}{2}}-1)$.
 We can use Euclid's algorithm to compute $\mathrm{gcd}(a^{\frac{r}{2}}+1, N)$ and $\mathrm{gcd}(a^{\frac{r}{2}}-1, N)$, which are factors of $N$.
 Thus, we find factors of $N$ provided $r$ is even and $a^{\frac{r}{2}} + 1 \not\equiv 0$ mod $N$.
@@ -532,7 +534,7 @@ \subsection{Shor's algorithm}
 $N = 15$ does not divide $50$, so $\gcd(50, N) = 5$ is a factor, and $\gcd(48, 15) = 3$ is a factor.
 \begin{theorem}
     Let $N$ be odd and not a prime power.
-    Then, choosing $a$ uniformly at random such that $\gcd(a,N) = 1$, the probability that $r$ is even and $(a^{\frac{r}{2}} + 1) \not\equiv 0$ mod $N$ is at least $\frac{1}{2}$.
+    Then, choosing $a$ uniformly at random s.t. $\gcd(a,N) = 1$, the probability that $r$ is even and $(a^{\frac{r}{2}} + 1) \not\equiv 0$ mod $N$ is at least $\frac{1}{2}$.
 \end{theorem}
 This implies that if $N$ is odd and not a prime power, we obtain a factor of $N$ with probability at least $\frac{1}{2}$.
 We repeat this process until the probability of not finding a factor is acceptably low.
@@ -549,18 +551,18 @@ \subsection{Shor's algorithm}
     \item Choose $1 < a < N$ uniformly at random and compute $b = \gcd(a,N)$.
     If $b > 1$, output $b$ and halt.
     \item Find the period $r$ of the modular exponentiation function $f(k) = a^k$ mod $N$.
-    If this fails, return to step (iii).
+    If this fails, return to step (3).
     \item If $r$ is even and $(a^{\frac{r}{2}} + 1) \not\equiv 0$ mod $N$, compute $t = \gcd(a^{\frac{r}{2}} + 1, N)$; if $1 < t < N$, output $t$ and halt.
-    Otherwise, return to step (iii).
+    Otherwise, return to step (3).
 \end{enumerate}
 We now describe the method to compute the period of the modular exponentiation function.
-Note that $f \colon \mathbb Z \to \mathbb Z$, not $\mathbb Z_N \to \mathbb Z_M$; we therefore cannot directly use the algorithm discussed previously.
+Note that $f \colon \mathbb Z \to \mathbb Z_N$, not $\mathbb Z_N \to \mathbb Z_M$; we therefore cannot directly use the algorithm discussed previously.
 We must first truncate the domain $\mathbb Z$ to some $\mathbb Z_M$.
 If $r$ is unknown, $f$ will not necessarily be periodic on $\mathbb Z_M$.
 However, if $M$ is $O(N^2)$, the single incomplete period has a negligible effect on the periodicity determination.
 We will define $M = 2^m$ for some $m$ and use $QFT_M$.
 
-Consider a finite domain $D = \qty{0, \dots, 2^m - 1}$, where $m$ is the smallest integer such that $2^m > N^2$.
+Consider a finite domain $D = \qty{0, \dots, 2^m - 1}$, where $m$ is the smallest integer s.t. $2^m > N^2$.
 Suppose $2^m = Br + b$ where $0 \leq b < r$, so $B = \floor*{\frac{2^m}{r}}$.
 We start with the equal superposition state $\ket{\psi_m} = \frac{1}{\sqrt{2^m}} \sum_{x \in D} \ket{x}$.
 Consider the quantum oracle $U_f$ corresponding to the modular exponentiation function $f$.
@@ -576,7 +578,7 @@ \subsection{Shor's algorithm}
 
 If $y = f(x_0)$ for $x_0 < b$, the probability of measuring $y$ is $\frac{B+1}{2^m}$.
 The post-measurement state of the first register is $\ket{\mathrm{per}} = \frac{1}{\sqrt{B+1}} \sum_{j=0}^B \ket{x_0 + jr}$.
-In the case $x_0 \geq b$, we have $\ket{\mathrm{per}} = \frac{1}{\sqrt{B}} \sum_{j=0}^{B-1} \ket{x_0 + jr}$.
+In the case $x_0 \geq b$, we have $\ket{\mathrm{per}} = \frac{1}{\sqrt{B}} \sum_{j=0}^{B-1} \ket{x_0 + jr}$ with prob $\frac{B}{2^m}$.
 In both cases,
 \[ \ket{\mathrm{per}} = \frac{1}{\sqrt{A}} \sum_{j=0}^{A-1} \ket{x_0 + jr} \]
 where $A = B+1$ if $y = f(x_0)$ with $x_0 < b$ and $A = B$ if $y = f(x_0)$ with $x_0 \geq b$.
@@ -586,32 +588,32 @@ \subsection{Shor's algorithm}
     &=  \frac{1}{\sqrt{A}} \frac{1}{\sqrt{2^n}} \sum_{c=0}^{2^m - 1} \omega^{x_0 c} \underbrace{\qty[\sum_{j = 0}^{A-1} (\omega^{cr})^j]}_{S} \ket{c} \\
 \end{align*}
 where $\omega = 2^{\frac{2\pi i}{M}}$ where $M = 2^m$.
-$S$ is a geometric series.
+$S$ is a geometric series with $\alpha = \omega^{cr}$.
 If $\frac{M}{r} \not\in \mathbb Z$, $\alpha^A \neq 1$.
 We claim that a measurement on $QFT_{2^m} \ket{\mathrm{per}}$ yields an integer $c$ which is close to a multiple of $\frac{M}{r}$ with high probability.
 
 Consider $k\frac{2^m}{r}$ for $k = 0, \dots, r-1$.
 Each of these multiples is within $\frac{1}{2}$ of a unique integer; indeed, $2^m = Br + b$ so $r < 2^m$, giving that $k\frac{2^m}{r}$ cannot be a half integer.
-Consider the values of $c$ such that $\abs{c - k \frac{2^m}{r}} < \frac{1}{2}$ for $k = 0, \dots, r-1$.
+Consider the values of $c$ s.t. $\abs{c - k \frac{2^m}{r}} < \frac{1}{2}$ for $k = 0, \dots, r-1$.
 % Note that $\omega^{cr} = 1$ if $e^{\frac{2\pi i cr}{M}} = 1$.
 \begin{theorem}
     Suppose that $QFT_{2^m} \ket{\mathrm{per}} = \sum_{c=0}^{2^m - 1} g(c) \ket{c}$, and that we measure the state and receive an outcome $c$.
-    Let $c_k$ be the unique integer such that $\abs{c_k - k\frac{2^m}{r}} < \frac{1}{2}$.
+    Let $c_k$ be the unique integer s.t. $\abs{c_k - k\frac{2^m}{r}} < \frac{1}{2}$.
     Then $\prob{c = c_k} > \frac{\gamma}{r}$ for a fixed constant $\gamma$ (which can be shown to be $\frac{4}{\pi^2}$).
     Moreover, the probability that $k, r$ are coprime is $\Omega\qty(\frac{1}{\log \log r})$ by the coprimality theorem.
 \end{theorem}
 Thus, with $O(\log \log N) > O(\log \log r)$ repetitions, we obtain a good $c$ value with high probability.
-Suppose that we measure $c$ such that $\abs{c - k \frac{2^m}{r}} < \frac{1}{2}$, so $\abs{\frac{c}{2^m} - \frac{k}{r}} < \frac{1}{2^{m+1}}$.
-Recall that $r < N$ and $m$ is minimal such that $2^m > N^2$.
+Suppose that we measure $c$ s.t. $\abs{c - k \frac{2^m}{r}} < \frac{1}{2}$, so $\abs{\frac{c}{2^m} - \frac{k}{r}} < \frac{1}{2^{m+1}}$.
+Recall that $r < N$ and $m$ is minimal s.t. $2^m > N^2$.
 Then $\abs{\frac{c}{2^m} - \frac{k}{r}} < \frac{1}{2N^2}$.
 Note that $\frac{c}{2^m}$ is known.
 
-We show that there is at most one fraction $\frac{k}{r}$ with denominator $r < N$ such that $\abs{\frac{c}{2^m} - \frac{k}{r}} < \frac{1}{2N^2}$.
+We show that there is at most one fraction $\frac{k}{r}$ with denominator $r < N$ s.t. $\abs{\frac{c}{2^m} - \frac{k}{r}} < \frac{1}{2N^2}$.
 Suppose $\frac{k'}{r'}, \frac{k''}{r''}$ both satisfy this requirement.
 Then
 \[ \abs{\frac{k'}{r'} - \frac{k''}{r''}} = \frac{\abs{k'r'' - k''r'}}{r'r''} \geq \frac{1}{r'r''} > \frac{1}{N^2} \]
 But $\abs{\frac{c}{2^m} - \frac{k'}{r'}}, \abs{\frac{c}{2^m} - \frac{k'}{r'}} < \frac{1}{2N^2}$, contradicting the triangle inequality.
-This result is the reason for choosing $m$ minimal such that $2^m > N^2$.
+This result is the reason for choosing $m$ minimal s.t. $2^m > N^2$.
 Therefore, we have with high probability that $\frac{c}{2^m}$ is close to a unique fraction $\frac{k}{r}$.
 \begin{example}
     Let $N = 39$ and choose $a = 7$; note that 7 and 39 are coprime.
@@ -659,12 +661,12 @@ \subsection{Shor's algorithm}
 \end{proof}
 \begin{theorem}
     Let $x \in \mathbb Q$ with $0 < x < 1$.
-    Let $\frac{p}{q} \in \mathbb Q$ such that $\abs{x - \frac{p}{q}} < \frac{1}{2q^2}$.
+    Let $\frac{p}{q} \in \mathbb Q$ s.t. $\abs{x - \frac{p}{q}} < \frac{1}{2q^2}$.
     Then $\frac{p}{q}$ is a convergent of the continued fraction expansion of $x$.
 \end{theorem}
-In our situation, we have $c$ such that
+In our situation, we have $c$ s.t.
 \[ \abs{\frac{c}{2^m} - \frac{k}{r}} < \frac{1}{2N^2};\quad r < N \]
-In particular, $\abs{\frac{c}{2^m} - \frac{k}{r}} < \frac{1}{2r^2}$, and we have seen that there is at most one fraction $\frac{k}{r}$ such that this holds.
+In particular, $\abs{\frac{c}{2^m} - \frac{k}{r}} < \frac{1}{2r^2}$, and we have seen that there is at most one fraction $\frac{k}{r}$ s.t. this holds.
 Note that $0 < c < 2^m$, so $0 < \frac{c}{2^m} < 1$.
 Hence, $\frac{k}{r}$ is a convergent of $\frac{c}{2^m}$.
 Note that $2^m > N^2 > 2^{m-1}$, so $c, 2^m$ are $O(m)$-bit integers, and hence the sequence of convergents (and in particular $\frac{k}{r}$) can be computed in $O(m^3)$ time.