Solved some exercises & Added those to Table Of Contents

tex/TheLuaChameleonBook.cls

@@ -76,6 +76,7 @@
 %----------------------------------------------------------------------------------------
 
 \usepackage{fontspec}
+\usepackage{lmodern} 
 \usepackage{ulem}
 
 %----------------------------------------------------------------------------------------
@@ -126,7 +127,7 @@
 		\node[anchor=north east, inner sep=0pt, xshift=-\Gm@rmargin, yshift=-\Gm@tmargin] at (current page.north east) {\fontsize{30pt}{30pt}\selectfont\sffamily\bfseries\textcolor{primarycolor}{\strut #2}}; % Part title
 		\node[anchor=south east, inner sep=0pt, xshift=-\Gm@rmargin, yshift=\Gm@bmargin] at (current page.south east) { % Mini table of contents
 			\parbox[t][][t]{8.5cm}{ % Width of box holding the mini ToC
-				\printcontents[part]{l}{0}{\setcounter{tocdepth}{1}} % Display the mini table of contents showing chapters and sections, change tocdepth to 2 to also show subsections or 0 to only show chapters
+				\printcontents[part]{l}{0}{\setcounter{tocdepth}{3}} % Display the mini table of contents showing chapters and sections, change tocdepth to 2 to also show subsections or 0 to only show chapters
 			}
 		};
 	\end{tikzpicture}

tex/content/chapters/include.tex

@@ -15,4 +15,4 @@
 \includechapters{part01}{2}
 
 
-\includechapters{part02}{2}
+% \includechapters{part02}{2}

tex/content/chapters/part01/0.tex

@@ -0,0 +1 @@
+\part{}

tex/content/chapters/part01/1.tex

@@ -0,0 +1,270 @@
+\chapter{Basic concepts}
+
+\section{Algorithms}
+
+\section{Mathematical Preliminaries}
+
+\subsection{Mathematical induction}
+
+\begin{myexercise}{4}{18}
+
+    Let $F_n$ be the $n$-th Fibonacci number, defined by the recursion relation
+    \[
+    \begin{cases}
+        F_0 = 1, \\
+        F_1 = 1, \\
+        F_n = F_{n-1} + F_{n-2}, \quad n \geq 2.
+    \end{cases}    
+    \]
+
+    Let $\phi = \frac{1+\sqrt{5}}{2}$ be the golden ratio. 
+
+    We have
+    \begin{equation}
+        F_n \leq \phi^{n-1}
+        \label{eq:fibo_leq_golden_ratio}
+    \end{equation}
+    $\forall n \in \mathbb{N}$.
+
+    Prove that, in addition to Eq. \ref{eq:fibo_leq_golden_ratio}, $F_n \geq \phi^{n-2}$.
+
+    \begin{myanswer}
+
+        Let $P(n)$ be the assertion that $F_n \geq \phi^{n-2}$.
+
+        We proceed by recursion on $n$.        
+        \begin{equation}
+            1 + \phi = \phi^2 
+            \label{eq:golden_ratio_squared}
+        \end{equation}
+
+        \emph{Initiation} For $n = 0$: $\phi^{0-2} = \phi^{-2} = \frac{1}{\phi^2}$. As $\phi^2 > 1$ . $\frac{1}{\phi^2} < 1$, so $P(0)$ is true.
+
+        For $n = 1$: 
+        $\phi^{1-2} = \phi^{-1} = \frac{1}{\phi}$. As $\phi > 1$, $\frac{1}{\phi} < 1$, so $P(1)$ is true.
+
+        \emph{Heredity}
+        Let $P(n)$ and $P(n-1)$ be true. We have to prove $P(n+1)$.
+        \begin{align*}
+            P(n) \wedge P(n-1) & \Leftrightarrow F_n \geq \phi^{n-2} \wedge F_{n-1} \geq \phi^{n-1-2}\\
+            & \Leftrightarrow F_{n} + F_{n-1} \geq \phi^{n-2} + \phi^{n-3} \\
+            & \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} (1 + \phi)  \\
+            & \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} \phi^2  \qquad \text{using Eq. \ref{eq:golden_ratio_squared}} \\
+            & \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-1} \\
+            & \Leftrightarrow F_{n+1} \geq \phi^{n+1-2} \\
+            & \Leftrightarrow P(n+1)
+        \end{align*}
+
+        \emph{Conclusion} $P(0)$ and $P(1)$ are true. $P(n) \wedge P(n-1)$ true implies $P(n+1)$ true. So $P(n)$ is true for all $n \in \mathbb{N}$.
+
+    \end{myanswer}
+\end{myexercise}
+
+
+
+\begin{myexercise}{5}{19}
+A prime number is an integer $> 1$ that has no exact divisors other than $1$ and itself.
+
+Using this definition and mathematical induction, prove that every integer $> 1$ can be written as a product of prime numbers or is a prime itself.
+
+\begin{myanswer}
+Let $P(n)$ be the assertion that $n$ can be written as a product of prime numbers or is a prime.
+\begin{proof}
+    
+    Let $N > 1$ be the smallest integer such that $P(N)$ is false.
+    
+    As $\forall N' < N$, $N' > 1$, $P(N')$ is true.
+    We have either of the following assertions:
+    
+    \begin{itemize}
+        \item $N$ is a prime number;
+        \item there exists $m$ and $n$ below $N$ such that, $m \times n = N$.
+        As $m$ and $n$ are below $N$, they satisfies $P$, so they are either primes or a product of primes. So $mn = N$ is also a product of primes.
+    \end{itemize}
+    We have a contradiction, and $P(N)$ must be true.
+    
+\end{proof}
+\end{myanswer}
+\end{myexercise}
+
+\begin{myexercise}{7}{19} 
+Formulate and prove by induction a rule for the sums $1^2$, $2^2 - 1^2$, $3^2 - 2^2 + 1^2$, $4^2 - 3^2 + 2^2 - 1^2$, $5^2 - 4^2 + 3^2 - 2^2 + 1^2$, etc. 
+
+\begin{myanswer}
+
+$1^2 = 1$
+
+$2^2 - 1^2 = 3$
+
+$3^2 - 2^2 + 1^2 = 6$
+
+$4^2 - 3^2 + 2^2 - 1^2 = 10$
+
+$5^2 - 4^2 + 3^2 - 2^2 + 1^2 = 15$
+
+
+We can rewrite the computed sequence with
+\[
+T_n = \sum_{i=0}^n (-1)^i (n-i)^2    
+\]
+
+With little help from formal computation, we get
+\begin{equation}
+T_n = \frac{1}{2} n (n + 1)
+\label{eq:ex7_half_factor_n_n+1}    
+\end{equation}
+
+\begin{proof}
+    Let us prove the formula. 
+
+    Let $P(n)$ be the proposition that Eq. \ref{eq:ex7_half_factor_n_n+1} is correct for $T_n$.
+
+    By recursion on $n$
+
+    \emph{Initiation}
+
+    For $n = 1$, $1^2 = 1$ and $\frac{1}{2} 1 \times (1+1) = 1$, so $P(n)$ is true.
+
+    \emph{Heredity}
+
+    Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
+    \begin{align*}
+        P(n) & \Leftrightarrow T_n = \frac{1}{2} n(n+1) \\
+        & \Leftrightarrow \sum_{i=0}^n (-1)^i (n-i)^2 = \frac{1}{2} n(n+1)
+    \end{align*}
+    \unfinished
+\end{proof}
+
+\end{myanswer}
+\end{myexercise}
+
+\begin{myexercise}{8}{19}
+    (a) Prove the following theorem of Nicomachus by induction:
+
+    $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$, $4^3 = 13 + 15 + 17 + 19$, etc.
+
+    (b) Use this result to prove the remarkable formula $1^3 + 2^3 +  \dots + n^3 = (1 + 2 + \dots + n)^2$
+
+    \begin{myanswer}
+
+    (a) The theorem states the following: 
+
+    For all $n \in \mathbb{N}$ we have
+    \[
+        n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|
+    \]
+
+    \begin{proof}
+        
+        We proceed by recursion on $n$.
+        
+        Let $P(n)$ be the proposition ``$n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|$''.
+        
+        \emph{Initiation} For $n = 1$, $1^3 = 1^2$, so $P(1)$ is true.
+        
+        \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
+        \begin{align*}
+            P(n) \Leftrightarrow n^3 &= \sum_{i = 1}^n n (n - 1) - 1 + 2i \\
+            % &\Leftrightarrow (n+1)^3 &= \sum_{i = 1}^n  n^2 - n - 1 + 2i \\
+        \end{align*}
+    \end{proof}
+    \unfinished
+
+    (b)
+
+    \begin{theorem}[Formula of Nicomachus]
+        The sum of the cubes of the first $n$ natural numbers is equal to the square of the sum of the first $n$ natural numbers.
+    \end{theorem}
+    \begin{equation}
+        \sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2
+    \end{equation}
+
+    \begin{proof}
+        Let $P(n)$ be the proposition ``$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$''.
+        
+        $P(n) \Longleftrightarrow \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$
+        because $\sum_{i=1}^n i = \frac{n(n+1)}{2}$.
+
+        We want to prove that: $\sum_{k=1}^{n+1} k^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$
+
+        \begin{align*}
+            \sum_{k=1}^{n+1} k^3 &= \frac{n^2 (n+1)^2}{2^2} + (n+1)^3 \\
+            &= (n+1)^2 \left(\frac{n^2}{4} + n + 1\right) \\
+            &= (n+1)^2 \cdot \frac{n^2 + 4n + 4}{4} \\
+            &= \frac{(n+1)^2(n+2)^2}{2^2} \\
+            & \Updownarrow \\
+            & P(n+1)
+        \end{align*}
+        
+        \emph{Conclusion}
+        
+        $P(1)$ is true because $1^3 = 1$ and $\left(\frac{1(1+1)}{2}\right)^2 = 1$.
+        
+        $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$.
+    \end{proof}
+    \end{myanswer}
+\end{myexercise}
+
+\begin{myexercise}{9}{19}
+    Prove by induction that if $0 < a < 1$ then $(1 - a)^n \geq 1 - na$.
+    \begin{myanswer}
+        We proceed by recursion on $n$.
+
+        Let $P(n)$ be the proposition ``$(1 - a)^n \geq 1 - na$''.
+
+        \emph{Initiation} For $n = 0$,
+
+        $(1 - a)^n = 1 = 1 - na$, so $P(0)$ is true.
+
+        \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
+
+        We want to prove that $(1 - a)^{n+1} \geq 1 - (n+1)a$.
+
+        \begin{align*}
+            P(n) \Leftrightarrow (1-a)^n &\geq 1-na \\
+            (1-a)^n (1-a) & \geq (1 - na) (1-a) \qquad \text{$1-a > 0 $} \\
+            (1-a)^{n+1} & \geq  1 - na - a + na \\ 
+            & \geq 1 - (n+1)a - a \\
+            & \geq 1 - (n+1)a \qquad \text{because $a > 0$}
+            & \Rightarrow P(n+1)
+        \end{align*}
+
+        \emph{Conclusion}
+        $P(1)$ is true. 
+
+        $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$.
+    \end{myanswer}
+\end{myexercise}
+
+% \begin{myexercise}{10}{19}
+%     Prove by induction that if $n \geq 10$ then $2^n > n^3$.
+
+%     \begin{myanswer}
+%         We proceed by recursion on $n$.
+
+%         Let $P(n)$ be the proposition ``$2^n > n^3$''.
+
+%         \emph{Initiation} For $n = 10$,
+
+%         $2^{10} = 1024 > 1000 = 10^3$, so $P(10)$ is true.
+
+%         \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
+
+%         We want to prove that $2^{n+1} > (n+1)^3$.
+
+%         \begin{align*}
+%             P(n) \Leftrightarrow 2^n &> n^3 \\
+%             2^n \times 2 &> n^3 \times 2 \qquad \text{because $2 > 0$} \\
+%             2^{n+1} &> n^3 + n^3 \\
+%         \end{align*}
+
+%         $n^3 = n^2 \times n$
+
+%         \emph{Conclusion}
+%         $P(10)$ is true. 
+
+%         $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all integer $n \geq 10$.
+%     \end{myanswer}
+% \end{myexercise}
+% Dirty false proof... To be continued
+

tex/content/introduction.tex

@@ -1,8 +1 @@
-\chapter{Introduction}
+\chapter*{Introduction}
-
-Here we go
-
-\part{Hello, world}
-
-
-\section{What did you expect ?}

tex/main.ist

@@ -1,5 +1,5 @@
 % makeindex style file created by the glossaries package
-% for document 'main' on 2022-12-31
+% for document 'main' on 2023-1-3
 actual '?'
 encap '|'
 level '!'

tex/main.tex

@@ -47,6 +47,9 @@
 \makeglossaries
 \makeindex
 
+\input{preamble.tex}
+\setcounter{tocdepth}{2}
+
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % DOCUMENT
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@@ -61,11 +64,11 @@
 	{\includegraphics[width=\paperwidth]{background.pdf}} % Code to output the background image, which should be the same dimensions as the paper to fill the page entirely; leave empty for no background image
 	{ % Title(s) and author(s)
 		\centering\sffamily % Font styling
-		{\Huge\bfseries Exploring the Physical Manifestation of Humanity's Subconscious Desires\par} % Book title
+		{\Huge\bfseries The Art of Computer Programming Vol. 1 - Some Exercises \par} % Book title
 		\vspace{16pt} % Vertical whitespace
-		{\LARGE A Practical Guide\par} % Subtitle
+		{\LARGE An Attempt to solve some of the exercises\\ from the Book of Donald E. KNUTH\par} % Subtitle
 		\vspace{24pt} % Vertical whitespace
-		{\huge\bfseries Goro Akechi\par} % Author name
+		{\huge\bfseries Samuel ORTION\par} % Author name
 	}
 
 

tex/preamble.tex

@@ -0,0 +1,66 @@
+
+\usepackage{xstring}
+
+\newcommand{\unfinished}{
+    \textbf{\textcolor{orange!90}{to be continued...}}
+}
+
+\newcommand{\unresolved}{
+    \textbf{\textcolor{red!90}{unresolved}}
+}
+
+% Create a new environment {exercise} to refer  to exercises from the KNUTH book.
+% Add a TODO switch option to show unresolved exercises
+\newenvironment{myexercise}[2]
+{
+    \addcontentsline{toc}{subsubsection}{Exercise #1}
+    \begin{minipage}{\textwidth}
+    \begin{flushleft}
+    {
+        \Huge
+        \textbf{Exercise #1.} (p. #2)
+    }
+    \end{flushleft}
+}
+{
+    \end{minipage}
+}
+
+% Create a new environment for big quote with `primarycolor' coloured quote sign
+\newenvironment{myquote}
+{
+    \begin{minipage}{\textwidth}
+    \begin{flushleft}
+    {
+
+        \fontsize{75pt}{3pt}\selectfont 
+        \textcolor{primarycolor}{\textquotedblleft}
+    }
+}
+{
+    \begin{flushright}
+       {
+           \textcolor{primarycolor}{\textquotedblright}
+       }
+
+    \end{flushright}
+    \end{flushleft}
+    \end{minipage}
+}
+
+\newenvironment{myanswer}
+{
+    \begin{minipage}{\textwidth}
+    \begin{itshape}
+    \begin{flushleft}
+    {
+        \Large
+        \textbf{My Answer}:
+    }
+}
+{
+    \end{flushleft}
+    \end{itshape}
+    \end{minipage}
+}
+