General corrections.

book/dixon.tex

@@ -1,15 +1,22 @@
-\chapter{Dixon \label{chap:dixon}}
+\chapter{Dixon {\texttt{\small{[insert random buzzword here]}}} method\label{chap:dixon}}
 
 ~\cite{dixon} describes a class of ``probabilistic algorithms'' for finding a
-factor of any composite number, at a computational cost asymptotically  best
-than all other ones previously described:
-\bigO{\beta(\log N \log \log N)^{\rfrac{1}{2}}}
-for some constant $\beta > 0$.
+factor of any composite number, at a sub-exponential cost. They basically
+consists into taking random integers $r$ in $\{1, \ldots, N\}$ and look for those
+where $r^2 \mod{N}$ is \emph{smooth}. If enough are found, then those integers
+can somehow be assembled, and so a fatorization of N attemped.
+
+
+\section{Quadratic Sieve}
+During the latest century there has been a huge effort to approach the problem
+formulated by Fermat ~\ref{eq:fermat_problem} from different perspecives. This
+led to an entire family of algorithms known as \emph{Quadratic Sieve} [QS]. The
+core idea is still to find a pair of perfect squares whose difference can
+factorize $N$, but maybe Fermat's hypotesis can be made weaker.
 
 \paragraph{Kraitchick} was the first one popularizing the idea the instead of
-looking for integers $\angular{x, y}$ such that $x^2 -y^2 = N$ -recall Fermat's
-problem, formulated in equation ~\ref{eq:fermat_problem}, it is sufficient to
-look for \emph{multiples} of $N$:
+looking for integers $\angular{x, y}$ such that $x^2 -y^2 = N$ it is sufficient
+to look for \emph{multiples} of $N$:
 \begin{align}
   x^2 - y^2 \equiv 0 \pmod{N}
 \end{align}
@@ -17,6 +24,13 @@ and, once found, claim that $\gcd(N, x \pm y)$ are non-trial divisors of $N$
 just as we did in \ref{sec:fermat:implementation}.
 On the top of this,
 
+\section{stuff}
+
+at a computational cost asymptotically  best
+than all other ones previously described:
+\bigO{\beta(\log N \log \log N)^{\rfrac{1}{2}}}
+for some constant $\beta > 0$.
+
 %%% Local Variables:
 %%% mode: latex
 %%% TeX-master: "question_authority"

book/fermat.tex

@@ -92,7 +92,7 @@ primes:
 
 \begin{bytefield}[
   endianness=big,
-  bitwidth=1.4em,
+  bitwidth=1.35em,
   % bitformatting=\fakerange,
   ]{16}
   \\

book/math_prequisites.tex

@@ -23,7 +23,8 @@ meaning of ``pick a uniformly distributed random element $x$ from the set $S$''.
 
 \section{Algorithmic Complexity Notation}
 The notation used to describe asymptotic complexity follows the $O$-notation,
-abused under the conventions and limits of MIT's Introduction to Algorithms.
+abused under the conventions and limits of MIT's Introduction to Algorithms
+\cite{MITalg}.
 
 Let \bigO{g} be the asymptotic upper bound of g:
 $$

book/pollardrho.tex

@@ -1,20 +1,20 @@
 \chapter{Pollard's $\rho$ factorization method \label{chap:pollardrho}}
 
-The \emph{Monte Carlo} factorization method, published by J. M. Pollard in
+A \emph{Monte Carlo} factorization method, published by J. M. Pollard in
 ~\cite{pollardMC}, consists into identifying a periodically recurrent  sequence
 of integers within a random walk $\pmod{N}$ that could leak one of the two
 factors.
 
 Consider a function $f$ from $\mathcal{S}$ to $\mathcal{S}$, where
-$\mathcal{S} = \{0, 1, \ldots, q-1\}$. Let $s$ be a random element in
-$\mathcal{S}$, and consider the sequence
+$\mathcal{S} = \{0, 1, \ldots, q-1\}$ and $q \in \naturalPrime$.
+Let $s$ be a random element in $\mathcal{S}$, and consider the sequence
 \begin{align*}
   s,\ f(s),\ f(f(s)),\ \ldots
 \end{align*}
 Since $f$ acts over a finite set, it is clear that this sequence must
 eventually repeat, and become cyclic.
 We might diagram it with the letter $\rho$, where the tail represent the
-cyclic part, or \emph{epacts}, and the oval the cyclic part, or
+aperiodic part, or \emph{epacts}, and the oval the cyclic part, or
 \emph{period}.
 \begin{center}
   \begin{tikzpicture}[scale=0.7, thick]
@@ -45,8 +45,8 @@ cyclic part, or \emph{epacts}, and the oval the cyclic part, or
 \end{center}
 
 Now, consider $N=pq$.
-Let $F(x)$ be any function generating pseudorandom numbers $\angular{x_1, x_2, \ldots}$,
-and let $f(x) = F(x) \pmod{q}$.
+Let $F(x)$ be any function generating pseudorandom integers
+$\angular{x_1, x_2, \ldots}$, and let $f(x) = F(x) \pmod{q}$.
 As we said above, without any luck, there will be a pair $\angular{x_i, x_j}$
 generated by $F$ such that $x_i \equiv x_j \pmod{q}$, but $x_i \neq x_j$.
 
@@ -58,18 +58,21 @@ factor of $N$.
 \paragraph{Choosing the function} Ideally, $F$ should be easily computable, but
 at the same time random enough to reduce as much as possible the epacts
 ~\cite{Crandall} \S 5.2.1. Any quadratic function $F(x) = x^2 + b$ should be
-enough, provided that $b \in \naturalN \setminus \{0, 2\}$ \footnote{
+enough \footnote{
   Note that this has been only empirically verified, and so far not been proved
-  (~\cite{riesel}, p. 177)}.
+  (~\cite{riesel}, p. 177)}
+, provided that $b \in \naturalN \setminus \{0, 2\}$.
 For example, ~\cite{pollardMC} uses $x^2 -1$, meanwhile we are going to choose
 $F(x) = x^2 + 1$.
 
-\paragraph{Finding the period} In \cite{AOCPv2} \S 3.1, Knuth gives a simple and
-elegant algorithm, attributed to Floyd, for finding a multiple of the
-period. This algorithm is the same one finally adopted by Pollard in
+\paragraph{Finding the period} The trivial way to discover a period would be to
+test $x_i$ with all $x_j, \quad j < i$. Though, in \cite{AOCPv2} \S 3.1,
+Knuth gives a simple and elegant algorithm, attributed to Floyd, for finding a
+multiple of the period.
+This algorithm is the same one finally adopted by Pollard in
 ~\cite{pollardMC}.
 
-\begin{theorem*}
+\begin{theorem*}[Floyd]
 Given an \emph{ultimately periodic} sequence, in the sense that there exists
 numbers $\lambda$ and $\mu$ for which the values:
 \begin{align*}
@@ -83,15 +86,15 @@ $\mu < n < \mu + \lambda$ such that $X_n = X_{2n}$.
 \begin{proof}
   First, if $X_n = X_{2n}$, then the sequence is obviously periodic from
   $X_{2n}$ onward, possibly even earlier.
-  Conversely, $X_n = X_m \quad (n \geq \mu)$ for
+  Conversely, it is true that $X_n = X_m \quad (n \geq \mu)$ for
   $m = n + k\lambda, \quad k \in \naturalN$. Hence, there will eventually
   be an $n$ such that $X_n = X_{2n}$ if and only if $n - \mu$ is a multiple of
   $\lambda$.
   The first such value happens for $n = (\lambda + 1)\floor{\rfrac{\mu}{\lambda}}$.
 \end{proof}
 
-The immediate consequence of this is that we can find the period $q$ simply by
-checking $\gcd(x_{2i} - x_i, N)$ for incremental $i$-s.
+The immediate consequence of this is that we can find a collision simply by
+checking $\gcd(x_{2i} - x_i, N) > 1$ for incremental $i$.
 
 \paragraph{Brent's Improvement} In 1979, Brent discovered an entire family of
 cycle-finding algorithms whose optimal version resulted to be 36\% faster than
@@ -100,7 +103,7 @@ Instead of looking for the period of the sequence using $x_{2i} - x_i$, Brent
 considers
 $\abs{x_j - x_{2^k}}$ for $ 3 \cdot 2^{k-1} < j \leq 2^{k+1}$, resulting in
 fewer operations required by the algorithm. Pragmatically, this boils down to
-confronting:
+compare:
 
 \medskip
 \begin{tabular}{l@{\hskip 40pt} l@{\hskip 50pt} l}
@@ -123,7 +126,7 @@ instead of Floyd's one runs around 25\% faster on average
 ~\cite{pollard-brent}.
 
 \section{Complexity}
-\cite{riesel} presents a nice demonstration of the \emph{average} complexity of
+\cite{riesel} presents a nice proof of the \emph{average} complexity of
 this algorithm, based on the birthday paradox.
 \newtheorem*{birthday}{The Birthday Paradox}
 \begin{birthday}
@@ -145,9 +148,9 @@ this algorithm, based on the birthday paradox.
   This expression becomes $< \rfrac{1}{2}$ for $\epsilon \geq 23$.
 \end{proof}
 
-We can obviously substitute the $365$ with any set cardinality $\zeta$
-to express the probability that a random function from $\integerZ_{|\epsilon}$
-to $\integerZ_{|\zeta}$ is injective. Back to our particular case,
+We can obviously substitute the $365$ with any set of cardinality $\zeta$
+to express the probability that a random function from $\integerZ_{\epsilon}$
+to $\integerZ_{\zeta}$ is injective. Back to our particular case,
 we want to answer the question:
 
 \emph{
@@ -190,7 +193,7 @@ and respectively updating them via $x \gets f(x)$ and $y \gets f(f(y))$.
 
 \begin{remark}
   It is intresting to see how in its basic version, Pollard's $\rho$
-  method just needs 3 variables are to preserve the
+  method just needs 3 variables  to preserve the
   state. This places it among the most parsimonious factorization algorithms in
   terms of memory footprint.
 \end{remark}

book/question_authority.tex

@@ -45,6 +45,7 @@
 \newcommand{\naturalN}{\mathbb{N}}
 \newcommand{\naturalPrime}{\mathbb{P}}
 \newcommand{\integerZ}{\mathbb{Z}}
+\newcommand{\factorBase}{\mathcal{B}}
 \newcommand{\bigO}[1]{\ensuremath{\operatorname{O}\left(#1\right)}}
 \newcommand{\openssl}{\textsc{OpenSSL}\ }
 %% here adopting Wikipedia's notation <https://en.wikipedia.org/wiki/Isqrt>

book/wiener.tex

@@ -27,7 +27,8 @@ a_0 + \frac{1}{a_1
     + \frac{1}{a_3
     + \frac{1}{a_4 + \ldots}}}}
 \end{align*}
-From now on, we eill consider only hereby described as a finite sequence of integers for convenience:
+Consider now any \emph{finite continued fraction}, conveniently represented with
+the sequence
 $\angular{a_0, a_1, a_2, a_3,  \ \ldots, a_n}$.
 Any number $x \in \mathbb{Q}$ can be represented as a finite continued fraction,
 and for each $i < n$ there exists a fraction $\rfrac{h_i}{k_i}$ approximating
@@ -54,9 +55,9 @@ By definition, each new approximation is recursively defined as:
   \end{cases}
 \end{align}
 
-After a small digression concerning the properties of continued fractions, Wiener, in
-~\cite{wiener}, shows that, if a continued fraction $f'$ is an underestimate of
-another one $f$, i.e.
+Among the prolific properties of such objects, firstly Wiener ~\cite{wiener}
+and later Boneh ~\cite{20years} discovered that, if a continued fraction $f'$ is
+an underestimate of another one $f$, i.e.
 \begin{align}
   f' = f(1-\delta)
 \end{align}
@@ -67,7 +68,7 @@ enough'', where small enough means:
   \delta = 1 - \frac{f'}{f} < \frac{1}{\rfrac{3}{2}{h_1}{k_1}}
 \end{align}
 \\
-The ``continued fraction algorithm'' allowing us to recover $f$ is the
+The \emph{continued fraction algorithm} allowing us to recover $f$ is the
 following:
 \begin{enumerate}[(i)]
   \setlength{\itemsep}{1pt}