diff --git a/.github/pull_request_template.md b/.github/pull_request_template.md
deleted file mode 100644
index 8a83e85f..00000000
--- a/.github/pull_request_template.md
+++ /dev/null
@@ -1,15 +0,0 @@
-# Prerequisite Checklist
-
-<!--
-Please mark items appropriately:
--->
-
-- [ ] Read the [contribution guidelines](https://github.com/open-education-hub/oer-template/blob/main/CONTRIBUTING.md#pull-requests) regarding submitting new changes to the project;
-- [ ] Tested your changes against relevant architectures and platforms;
-- [ ] Updated relevant documentation (if needed).
-
-## Description of changes
-
-<!--
-Please provide a detailed description of the changes made in this new PR.
--->
diff --git a/.github/workflows/actions.yml b/.github/workflows/actions.yml
deleted file mode 100644
index 9e25a1f3..00000000
--- a/.github/workflows/actions.yml
+++ /dev/null
@@ -1,34 +0,0 @@
-name: Linter Actions Workflow
-
-on:
-  pull_request:
-    branches:
-      - main
-
-jobs:
-  checkpatch:
-    name: Checkpatch
-    runs-on: ubuntu-latest
-    steps:
-      - name: Checkpatch
-        uses: open-education-hub/actions/checkpatch@main
-        with:
-          GITHUB_TOKEN: ${{ secrets.GITHUB_TOKEN }}
-
-  super-linter:
-    name: Super Linter
-    runs-on: ubuntu-latest
-    steps:
-      - name: Super Linter
-        uses: open-education-hub/actions/super-linter@main
-        with:
-          GITHUB_TOKEN: ${{ secrets.GITHUB_TOKEN }}
-
-  spellcheck:
-    name: Spellcheck
-    runs-on: ubuntu-latest
-    steps:
-      - name: Spellcheck
-        uses: open-education-hub/actions/spellcheck@main
-        with:
-          GITHUB_TOKEN: ${{ secrets.GITHUB_TOKEN }}
diff --git a/slides/courses/1/components.tex b/slides/courses/1/components.tex
index 0ffeb0d7..7b50ab26 100644
--- a/slides/courses/1/components.tex
+++ b/slides/courses/1/components.tex
@@ -95,5 +95,5 @@
         \centering
         \scalebox{0.8}{\usebox{\asciiplcomp}}
     \end{figure}
-    Rolul limbajelor de programare este de a crea iluzia unei ierarhi de calculatoare virtuale.
-\end{frame}
\ No newline at end of file
+    Rolul limbajelor de programare este de a crea iluzia unei ierarhii de calculatoare virtuale.
+\end{frame}
diff --git a/slides/courses/1/functional.tex b/slides/courses/1/functional.tex
index 39c43717..21bf0915 100644
--- a/slides/courses/1/functional.tex
+++ b/slides/courses/1/functional.tex
@@ -1,7 +1,7 @@
 \begin{frame}
     \frametitle{Niveluri funcționale ale unui CN}
 \begin{enumerate}
-    \item Dispozitve și circuite electronice (Transiztoare, portți logice) - hardware
+    \item Dispozitve și circuite electronice (Tranzistoare, porți logice) - hardware
     \item Unități funcționale (UAL, Memorie, Interfețe) - firmware (Micropgramare) + hardware
     \item Mașină fizică Strcuturi de interconectare și echipamente periferice - hardware
     \item Nucleul sistem de operare - BIOS - firmware
@@ -28,4 +28,4 @@
     \item Tendința actuală este de a trece cât mai multe funcții în hardware
 \end{itemize}
 
-\end{frame}
\ No newline at end of file
+\end{frame}
diff --git a/slides/courses/2/intro.tex b/slides/courses/2/intro.tex
new file mode 100644
index 00000000..ce548304
--- /dev/null
+++ b/slides/courses/2/intro.tex
@@ -0,0 +1,79 @@
+\begin{frame}
+    \frametitle{Information Representation}
+    \begin{itemize}
+        \item Decimal
+        \item Binary
+        \item Hexadecimal
+    \end{itemize}
+    \note{
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Information Representation}
+
+    \begin{table}[h!]
+        \centering
+        \scalebox{0.8}{
+        \begin{tabular}{|c|c|c|}
+            \hline
+            \textbf{Decimal} & \textbf{Binary} & \textbf{Hexadecimal} \\
+            \hline
+            0 & 0000 & 0x0 \\
+            1 & 0001 & 0x1 \\
+            2 & 0010 & 0x2 \\
+            3 & 0011 & 0x3 \\
+            4 & 0100 & 0x4 \\
+            5 & 0101 & 0x5 \\
+            6 & 0110 & 0x6 \\
+            7 & 0111 & 0x7 \\
+            8 & 1000 & 0x8 \\
+            9 & 1001 & 0x9 \\
+            10 & 1010 & 0xA \\
+            11 & 1011 & 0xB \\
+            12 & 1100 & 0xC \\
+            13 & 1101 & 0xD \\
+            14 & 1110 & 0xE \\
+            15 & 1111 & 0xF \\
+            \hline
+        \end{tabular}
+        }
+        \caption{Decimal, Binary, and Hexadecimal Values}
+        \label{tab:binary_decimal_hexadecimal}
+    \end{table}
+
+    \note{
+    }
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Information Representation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Big Endian vs Little Endian}
+    \begin{itemize}
+        \item Big Endian
+        \begin{itemize}
+            \item Most significant byte first
+            \item Network byte order
+            \item Example: 0x12345678 is stored as 0x12 0x34 0x56 0x78
+        \end{itemize}
+        \item Little Endian
+        \begin{itemize}
+            \item Least significant byte first
+            \item Intel byte order
+            \item Example: 0x12345678 is stored as 0x78 0x56 0x34 0x12
+        \end{itemize}
+    \end{itemize}
+\end{frame}
+
+
+
+
+\begin{frame}
+    \frametitle{Exam Question}
+    Given the following hexadecimal number $0x3A2E$ in big endian format, convert it to binary and decimal.
+    $0x3A2E = 0011\ 1010\ \ 0010\ 1110 = (3 * 16 + 10) * 2^8 + (2 * 16 + 14) = 58 * 256 + 46 = 14894$
+\end{frame}
diff --git a/slides/courses/2/main.tex b/slides/courses/2/main.tex
new file mode 100644
index 00000000..0a94ae14
--- /dev/null
+++ b/slides/courses/2/main.tex
@@ -0,0 +1,113 @@
+%\documentclass[notes,usenames,dvipsnames]{beamer}
+%\documentclass[notes]{beamer}       % print frame + notes
+%\documentclass[notes=only]{beamer}   % only notes
+\documentclass[usenames,dvipsnames]{beamer}             % only frames
+\usepackage{pgfpages}
+%\setbeameroption{show notes on second screen}
+%\setbeameroption{show notes}
+%subfigures
+\usepackage{caption}
+\usepackage{subcaption}
+
+%tables packages
+\usepackage{multirow}
+
+% math
+\usepackage{amsmath}
+
+% bash command
+\usepackage{graphicx}
+\usepackage{listings}
+% varbatim for ascii figures
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+%\usepackage{verbatim}
+\usepackage{lipsum} % for context
+\usepackage{fancyvrb}
+\usepackage{varwidth}
+
+
+\newsavebox{\asciigcn}
+
+% notes prefixed in pympress
+\addtobeamertemplate{note page}{}{\thispdfpagelabel{notes:\insertframenumber}}
+%theme used
+\usetheme{Madrid}
+%Information to be included in the title page:
+\title[Computer Architecture] %optional
+{Computer Architecture}
+
+\subtitle{Course no. 2 - Information Representation}
+
+\author[Ștefan-Dan Ciocîrlan] % (optional, for multiple authors)
+{}
+
+\institute[NUSTPB] % (optional)
+{
+  \inst{}%
+  National University of Science and Technology\\
+  POLITEHNICA Bucharest
+}
+
+\date[NUSTPB 2024] % (optional)
+{Computer Architecture}
+
+\logo{
+\includegraphics[height=0.9cm]{../../media/LOGO_UNSTPB_en.png}
+\includegraphics[height=0.9cm]{../../media/logoACSQ.jpeg}
+}
+
+
+% Roman numerals
+\newcommand*{\rom}[1]{\expandafter\@slowromancap\romannumeral #1@}
+%split
+\usepackage{amsmath}
+%colors
+%\usepackage[usenames,dvipsnames]{color} %loaded by the dcoument class
+%subfigure
+\usepackage{subcaption}
+% block over block uncover
+\setbeamercovered{invisible}
+%\setbeamercovered{transparent}
+
+%extra slide content
+\AtBeginSection[]
+{
+  \begin{frame}
+    \frametitle{Content}
+    \tableofcontents[currentsection]
+  \end{frame}
+}
+%notes or not
+%\setbeamertemplate{note page}[plain]
+
+\begin{document}
+
+\frame{\titlepage}
+
+\section{Information Representation}
+\input{intro.tex}
+
+\section{Textual Representation Systems}
+\input{text.tex}
+
+\section{Number Representation Systems}
+\input{number.tex}
+
+%\section{Image Representation Systems}
+%\input{image.tex}
+
+%\section{Audio Representation Systems}
+%\input{audio.tex}
+
+\section{Q\&A}
+\begin{frame}
+\end{frame}
+
+%\begin{frame}
+%\frametitle{Table of Contents}
+%\tableofcontents
+%\end{frame}
+
+
+\end{document}
\ No newline at end of file
diff --git a/slides/courses/2/media/International_Morse_Code.png b/slides/courses/2/media/International_Morse_Code.png
new file mode 100644
index 00000000..2cf3c590
Binary files /dev/null and b/slides/courses/2/media/International_Morse_Code.png differ
diff --git a/slides/courses/2/media/Morse_code_tree3.png b/slides/courses/2/media/Morse_code_tree3.png
new file mode 100644
index 00000000..8321065a
Binary files /dev/null and b/slides/courses/2/media/Morse_code_tree3.png differ
diff --git a/slides/courses/2/media/add.jpg b/slides/courses/2/media/add.jpg
new file mode 100644
index 00000000..90c1d192
Binary files /dev/null and b/slides/courses/2/media/add.jpg differ
diff --git a/slides/courses/2/media/mul.jpg b/slides/courses/2/media/mul.jpg
new file mode 100644
index 00000000..8b4af37e
Binary files /dev/null and b/slides/courses/2/media/mul.jpg differ
diff --git a/slides/courses/2/number.tex b/slides/courses/2/number.tex
new file mode 100644
index 00000000..5a67c8cc
--- /dev/null
+++ b/slides/courses/2/number.tex
@@ -0,0 +1,434 @@
+\subsection{Integer}
+
+\begin{frame}
+    \frametitle{Unary}
+    Number of consecutive 1 bits with 0 as the last bit
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{black}{b\textsubscript{n-1}b\textsubscript{n-2}...b\textsubscript{1}b\textsubscript{0}}\\
+            &b\textsubscript{i}=
+                \begin{cases}
+                    1,& \text{if } x\geq i+1\\
+                    0, & \text{otherwise}
+                \end{cases}
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Binary}
+    The number representation in base 2
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{black}{b\textsubscript{n-1}b\textsubscript{n-2}...b\textsubscript{1}b\textsubscript{0}}\\
+            &x=\sum_{i=0}^{n-1} b_i \times 2^i
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Sign magnitude}
+    The binary number representation system with a sign bit
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{red}{s}\textcolor{black}{b\textsubscript{n-2}b\textsubscript{n-3}...b\textsubscript{1}b\textsubscript{0}}\\
+            &x=(-1)^{\textcolor{red}{s}} \times \sum_{i=0}^{n-2} b_i \times 2^i
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{One's complement}
+    The bynary number representation with one's complement
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{red}{s}\textcolor{black}{b\textsubscript{n-2}b\textsubscript{n-3}...b\textsubscript{1}b\textsubscript{0}}\\
+            &x=\begin{cases}
+                    \sum_{i=0}^{n-2} b_i \times 2^i,& \text{if \textcolor{red}{sign} positive} \\
+                    (-1) \times \sum_{i=0}^{n-2} \overline{b_i} \times 2^i, & \text{otherwise}
+            \end{cases} 
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Two's complement}
+    The binary number representation with two's complement
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{red}{s}\textcolor{black}{b\textsubscript{n-2}b\textsubscript{n-3}...b\textsubscript{1}b\textsubscript{0}}\\
+            &x=\begin{cases}
+                    \sum_{i=0}^{n-2} b_i \times 2^i,& \text{if \textcolor{red}{sign} positive} \\
+                    (-1) \times ((\sum_{i=0}^{n-2} \overline{b_i} \times 2^i) + 1), & \text{otherwise}
+            \end{cases} 
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Bias}
+    The binary number representation with bias
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{black}{b\textsubscript{n-1}b\textsubscript{n-2}...b\textsubscript{1}b\textsubscript{0}}\\
+            &\text{Bias}=2^{n-1}-1\\
+            &x=(\sum_{i=0}^{n-1} b_i \times 2^i) - \text{Bias}
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Unary negative}
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{black}{b\textsubscript{n-1}b\textsubscript{n-2}...b\textsubscript{1}b\textsubscript{0}}\\
+            &b\textsubscript{i}=
+                \begin{cases}
+                    1, & \text{if } (x\geq i \text{ and } x \geq 0)  \text{ or } (|x| \leq i \text{ and } x < 0)\\
+                    0,& \text{otherwise}\\
+                \end{cases}
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Unary}
+\end{frame}
+
+\subsection{Fractional}
+\begin{frame}
+    \frametitle{Fractional}
+    Fractional numbers as $Q(ns, ds)$, where $ns$ is the numerator size and $ds$ is the denominator size.
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{red}{s}\textcolor{black}{n\textsubscript{ns-2}n\textsubscript{ns-3}...n\textsubscript{1}n\textsubscript{0}}\textcolor{orange}{d\textsubscript{ds-1}d\textsubscript{ds-2}...d\textsubscript{1}d\textsubscript{0}}\\
+            &x=(-1)^{\textcolor{red}{s}} \times \frac{\sum_{i=0}^{ns-2} n_i \times 2^i}{\textcolor{orange}{\sum_{i=0}^{ds-1} d_i \times 2^i}}
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Fractional}
+\end{frame}
+
+\subsection{Fixed Point}
+\begin{frame}
+    \frametitle{Fixed Point}
+    Fixed point numbers as $FP(is, fs)$, where $is$ is the integer part size and $fs$ is the fractional part size.
+    \begin{equation}
+        \begin{aligned}
+            &\text{Binary}\textsubscript{x}=\textcolor{red}{s}\textcolor{black}{i\textsubscript{is-2}i\textsubscript{is-3}...i\textsubscript{1}i\textsubscript{0}}\textcolor{orange}{f\textsubscript{fs-1}f\textsubscript{fs-2}...f\textsubscript{1}f\textsubscript{0}}\\
+            &x=\begin{cases}
+                    (\sum_{j=0}^{is-2} i_j \times 2^j) + \textcolor{orange}{(\sum_{j=0}^{fs-1} f_j \times 2^{j-fs})},& \text{if \textcolor{red}{sign} positive} \\
+                    (-1) \times ((\sum_{j=0}^{is-2} \overline{i}_j \times 2^j) + \textcolor{orange}{(\sum_{j=0}^{fs-1} \overline{f}_j \times 2^{j-fs})} + \frac{1}{ \textcolor{orange}{2^{fs}}}), & \text{otherwise}
+            \end{cases}
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Fixed Point}
+\end{frame}
+
+\subsection{Floating Point}
+\begin{frame}
+    \frametitle{Simple Floating Point}
+    Simple floating point numbers as $FloatP(es, fs)$, where $es$ is the exponent size and $fs$ is the fractional part size (mantissa).
+    $x=(-1)^{\text{sign}} \times 2^{\text{exponent}}  \times 1.f$.
+    \begin{equation}
+        \begin{aligned}
+            \text{Binary}\textsubscript{x}&=\textcolor{red}{s}\textcolor{ForestGreen}{e\textsubscript{es-1}e\textsubscript{es-2}...e\textsubscript{1}e\textsubscript{0}}\textcolor{orange}{m\textsubscript{fs-1}m\textsubscript{fs-2}...m\textsubscript{1}m\textsubscript{0}}\\
+            \text{Bias}&=2^{es-1}-1\\
+            e&=\textcolor{ForestGreen}{\sum_{i=0}^{es-1} e_i \times 2^{i}} - \text{Bias}\\
+            x&=(-1)^{\textcolor{red}{s}} \times \begin{cases}
+                    0, \text{if } \forall\ i<es\ e_i=0\ \And \forall\ i<fs\ m_i=0 \\
+                    2^e \times (1 + \textcolor{orange}{\frac{\sum_{i=0}^{fs-1} m_i \times 2^i}{2^{fs}}}), \text{otherwise}
+            \end{cases}
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Simple Floating Point}
+\end{frame}
+
+\begin{frame}
+    \frametitle{IEEE754 Floating Point}
+    Can use the notation $IEEE754(es, fs)$, where $es$ is the exponent size and $fs$ is the fractional part size (mantissa).
+\begin{equation}
+    \resizebox{0.9\textwidth}{!}{%
+    $
+    \begin{aligned}
+        x&=(-1)^{\textcolor{red}{s}} \times \begin{cases}
+                0 &, \text{if } \textcolor{ForestGreen}{\forall\ i<es\ e_i=0}\ \And \textcolor{orange}{\forall\ i<fs\ m_i=0} \\
+                2^{\textcolor{ForestGreen}{e}+1} \times \textcolor{orange}{\frac{\sum_{i=0}^{fs-1} m_i \times 2^i}{2^{fs}}} &, \text{if } \textcolor{ForestGreen}{\forall\ i<es\ e_i=0}\ \And \textcolor{orange}{\exists\ i<fs\ m_i\neq0} \\
+                \textcolor{Plum}{\infty} &, \text{if } \textcolor{ForestGreen}{\forall\ i<es\ e_i=1}\ \And \textcolor{orange}{\forall\ i<fs\ m_i=0} \\
+                 \textcolor{Plum}{sNaN} &, \text{if } \textcolor{ForestGreen}{\forall\ i<es\ e_i=1}\ \And \textcolor{orange}{m_{fs-1}=0} \And \textcolor{orange}{\exists\ i<fs-1\ m_i\neq0} \\
+                 \textcolor{Plum}{qNaN} &, \text{if } \textcolor{ForestGreen}{\forall\ i<es\ e_i=1}\ \And \textcolor{orange}{m_{fs-1}=1} \\
+                2^{\textcolor{ForestGreen}{e}} \times (1 + \textcolor{orange}{\frac{\sum_{i=0}^{fs-1} m_i \times 2^i}{2^{fs}}}) &, \text{otherwise}
+        \end{cases}
+    \end{aligned}$%
+    }
+\end{equation}
+
+\end{frame}
+
+\begin{frame}
+    \frametitle{IEEE754 Floating Point}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Morris Floating Point}
+    In the Morris floating point representation we use the following notation $Morris(s, g)$, where $s$ is size and $g$ is the size of the exponent size.
+    \begin{equation}
+        \resizebox{0.9\textwidth}{!}{%
+        $
+        \begin{aligned}
+            \text{Binary}\textsubscript{x}&=\textcolor{blue}{G\textsubscript{g-1}G\textsubscript{g-2}...G\textsubscript{1}G\textsubscript{0}}\textcolor{red}{s\textsubscript{e}}\textcolor{ForestGreen}{e\textsubscript{es-1}e\textsubscript{es-2}...e\textsubscript{1}e\textsubscript{0}}\textcolor{red}{s\textsubscript{f}}\textcolor{orange}{m\textsubscript{fs-1}m\textsubscript{fs-2}...m\textsubscript{1}m\textsubscript{0}}\\
+            G&=\textcolor{blue}{\sum_{i=0}^{g-1} G_i \times 2^{i}}\\
+            es&=G+1\\
+            e&=(-1)^{\textcolor{red}{s\textsubscript{e}}}\textcolor{ForestGreen}{\sum_{i=0}^{es-1} e_i \times 2^{i}}\\
+            x&=(-1)^{\textcolor{red}{s\textsubscript{f}}} \times \begin{cases}
+                    0 &, \text{if } \forall\ i\ \textcolor{ForestGreen}{e_i=0}\ \And\ \textcolor{orange}{m_i=0}\ \And\ \textcolor{blue}{G_i=0}\\
+                    NR &, \text{all bits $1$}\\
+                    2^{\textcolor{ForestGreen}{e}} \times (1 + \textcolor{orange}{\frac{\sum_{i=0}^{fs-1} m_i \times 2^i}{2^{fs}}}) &, \text{otherwise}
+            \end{cases}
+        \end{aligned}$%
+        }
+    \end{equation}
+    \note{
+   The Morris floating point system is given by the size and the size of the exponent size called simple $g$. The first $g$ bits represent the value $G$, which is added with $1$ to obtain the exponent size ($es$). The next bit is the exponent sign bit, followed by $es$ bits, which represent the absolute value of the exponent. The next bit is the fraction sign, and the remaining bits are considered fraction bits. The hidden bit is always $1$. The value is computed by $2^{\text{exponent}} \times (-1)^{\text{fractionSign}} \times (1+\frac{f}{2^{fs}})$, where exponent is $(-1)^{\text{exponentSign}} \times \text{exponentAbsoluteValue}$. The $0$ is represented by all bits of zero, and error cases are represented by only $1s$
+    }
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Morris Floating Point}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Posit}
+    $Posit(size, es)$ is given by $size$ and exponent size ($es$).
+    \begin{equation}
+        \resizebox{0.9\textwidth}{!}{%
+        $
+        \begin{aligned}
+            \text{Binary}\textsubscript{x}&=\textcolor{red}{s}\textcolor{black}{b\textsubscript{size-2}b\textsubscript{size-3}...b\textsubscript{1}b\textsubscript{0}}\\
+            \text{AbsoluteBits}\textsubscript{x}&=\begin{cases}
+                \overline{\text{Binary}\textsubscript{x}} + 1 &, \text{if \textcolor{red}{sign} negative}\\
+                 \text{Binary}\textsubscript{x} &, \text{otherwise}
+            \end{cases}\\
+            \text{AbsoluteBits}\textsubscript{x}&=\textcolor{red}{0}\textcolor{blue}{r\textsubscript{0}r\textsubscript{1}...r\textsubscript{rs-2}\overline{r\textsubscript{rs-1}}}\textcolor{ForestGreen}{e\textsubscript{es-1}e\textsubscript{es-2}...e\textsubscript{1}e\textsubscript{0}}\textcolor{orange}{m\textsubscript{fs-1}m\textsubscript{fs-2}...m\textsubscript{1}m\textsubscript{0}}\\
+            k&=\begin{cases}
+                \textcolor{blue}{rs}-2 &, \text{if } \textcolor{blue}{r_0=1}\\
+                -(\textcolor{blue}{rs}-1) &, \text{otherwise}
+            \end{cases}\\
+            e&=\textcolor{blue}{k} \times 2^{es} + \textcolor{ForestGreen}{\sum_{i=0}^{es-1} e_i \times 2^{i}}\\
+            x&=(-1)^{\textcolor{red}{s}} \times \begin{cases}
+                    0 &, \text{all bits of \text{Binary}\textsubscript{x} $0$}\\
+                    NaR &, \text{first bit $1$ and the rest $0$ for \text{Binary}\textsubscript{x}}\\
+                    2^{\textcolor{ForestGreen}{e}} \times (1 + \textcolor{orange}{\frac{\sum_{i=0}^{fs-1} m_i \times 2^i}{2^{fs}}}) &, \text{otherwise}
+            \end{cases}
+        \end{aligned}$%
+        }
+    \end{equation}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Posit}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Until now}
+    \begin{table}[ht]
+        %\begin{center}
+        \resizebox{0.65\columnwidth}{!}{%
+        \begin{tabular}{|p{1em}|p{5.7em}|p{3.4em}|p{16em}|c|c|c|c|}
+            \hline
+            \multirow{2}{*}{Set} & \multirow{2}{*}{NRS} & \multirow{2}{*}{Precision} & \multirow{2}{*}{m-layer} &  \multirow{2}{*}{$[x]$} & \multicolumn{3}{|c|}{s-layer given  h-layer} \\
+            \cline{6-8}
+            &&&&&5&-5&1.25\\
+            \hline
+            % N natural numbers set
+            %\iffalse
+            \multirow{4}{*}{$\mathbb{N}$} & \multirow{2}{*}{Unary} & IP & $\mathbb{N}$ & & 11111 & NR & NR\\
+            \cline{3-8}
+            & & VP/FP& $\{x, x \in \mathbb{N} , x < n \}$ & $U[8,r]$ & 00011111 & NR & NR\\
+            \cline{2-8}
+            & \multirow{2}{*}{Base 2} & IP & $\mathbb{N}$ & & 101 & NR & NR\\
+            \cline{3-8}
+            & & VP/FP & $\{x, x \in \mathbb{N} , x < 2^n \}$ & $\mathbb{N}[8,r]$ & 00000101 & NR & NR\\
+            \hline
+            % Z integer numbers set
+            \multirow{8}{*}{$\mathbb{Z}$} & \multirow{2}{*}{Sign} & IP & $\mathbb{Z}$ & & \textcolor{red}{0}101 & \textcolor{red}{1}101 & NR\\
+            \cline{3-8}
+            & & VP/FP & $\{(s,x), s \in \{0,1\}, x \in \mathbb{N}[n-1, r] \}$ & $\mathbb{Z}[8,r]$ & \textcolor{red}{0}0000101 & \textcolor{red}{1}0000101 & NR\\
+            \cline{2-8}
+            & \multirow{2}{*}{1'complement} & IP & $\mathbb{Z}$ & & \textcolor{red}{0}101 & \textcolor{red}{1}010 & NR\\
+            \cline{3-8}
+            & & VP/FP & $\{x, x \in \mathbb{Z} , |x| < 2^n \}$ & [8,r] & \textcolor{red}{0}0000101 & \textcolor{red}{1}1111010 & NR\\
+            \cline{2-8}
+            & \multirow{2}{*}{2'complement} & IP & $\mathbb{Z}$ & & \textcolor{red}{0}101 & \textcolor{red}{1}011 & NR\\
+            \cline{3-8}
+            & & VP/FP & $\{x, x \in \mathbb{Z} , -2^{n-1} \leq x < 2^{n-1} \}$ & [8,r] & \textcolor{red}{0}0000101 & \textcolor{red}{1}1111011 & NR\\
+            \cline{2-8}
+            &\multirow{2}{*}{Bias} & IP & $\{y-bias, y \in \mathbb{N}\}$ & [10] & 1111 & 0101 & NR\\
+            \cline{3-8}
+            & & VP/FP & $\{y-bias, y \in \mathbb{N}[n, r]\}$ & $\text{Bias}[10,8,r]$ & 00001111 & 00000101 & NR\\
+            \hline
+            % Q rational numbers set
+            \multirow{12}{*}{$\mathbb{Q}$} & \multirow{2}{*}{Fractional} & IP & $\mathbb{Q}$ & & \textcolor{red}{0}101/1 & \textcolor{red}{1}101/1 & \textcolor{red}{0}101/100\\
+            \cline{3-8}
+            & & VP/FP & $\{\frac{x}{y}, x \in \mathbb{Z}[ns, r], y \in \mathbb{N}[ds, r]  \}$ & $\mathbb{Q}[4,3,r]$ & \textcolor{red}{0}101\textcolor{orange}{001} & \textcolor{red}{1}101\textcolor{orange}{001} & \textcolor{red}{0}101\textcolor{orange}{100}\\
+            \cline{2-8}
+            %FixedPoint
+            & \multirow{2}{*}{Fixed Point} & IP & $\{\frac{x}{2^y}, x \in \mathbb{Z}, y \in \mathbb{N}  \}$ & & \textcolor{red}{0}101\textcolor{blue}{.}0 & \textcolor{red}{1}101\textcolor{blue}{.}0 & \textcolor{red}{0}001\textcolor{blue}{.}10\\
+            \cline{3-8}
+            & & VP/FP & $\{\frac{x}{2^{bp}}, x \in \mathbb{Z}[is+fs, r]  \}$ & $\text{FixedPoint}[4,3,r]$ & \textcolor{red}{0}101\textcolor{blue}{.}\textcolor{orange}{000} & \textcolor{red}{1}101\textcolor{blue}{.}\textcolor{orange}{000} & \textcolor{red}{0}001\textcolor{blue}{.}\textcolor{orange}{100}\\
+            \cline{2-8}
+            % Floating Point
+            & \multirow{2}{*}{Floating Point} & IP & $\{x \times 2^y, x \in [1,2), y \in \mathbb{Z}  \}$ & $\text{FloatingPoint}[\infty]$ & \textcolor{red}{0}1\textcolor{blue}{.}\textcolor{orange}{010}/\textcolor{red}{0}\textcolor{ForestGreen}{10} & \textcolor{red}{1}1\textcolor{blue}{.}\textcolor{orange}{010}/\textcolor{red}{0}\textcolor{ForestGreen}{10} & \textcolor{red}{0}1\textcolor{blue}{.}\textcolor{orange}{100}/\textcolor{red}{0}\textcolor{ForestGreen}{0}\\
+            \cline{3-8}
+            & & VP/FP & $\{(1 + \frac{f}{2^{fs}}) \times 2^{exp}, (f,exp) \in M, M = ((\mathbb{N}[fs, r] \times \text{Bias}[es, 2^{es-1}-1, r]) \setminus \{(0, -2^{es-1}+1)\})  \} \cup \{0\}$ & \text{FloatingPoint}[4,3,r] & \textcolor{red}{0}\textcolor{ForestGreen}{1001}\textcolor{orange}{010} & \textcolor{red}{1}\textcolor{ForestGreen}{1001}\textcolor{orange}{010} & \textcolor{red}{0}\textcolor{ForestGreen}{0111}\textcolor{orange}{100}\\
+            \cline{2-8}
+            %IEEE754
+            & \multirow{2}{*}{IEEE754} & IP & $\text{FloatingPoint}[\infty]$ & & \textcolor{red}{0}1\textcolor{blue}{.}\textcolor{orange}{010}/\textcolor{red}{0}\textcolor{ForestGreen}{10} & \textcolor{red}{1}1\textcolor{blue}{.}\textcolor{orange}{010}/\textcolor{red}{0}\textcolor{ForestGreen}{10} & \textcolor{red}{0}1\textcolor{blue}{.}\textcolor{orange}{100}/\textcolor{red}{0}\textcolor{ForestGreen}{0}\\
+            \cline{3-8}
+            & & VP/FP & $\{(1 + \frac{f}{2^{fs}}) \times 2^{exp}, x \in \mathbb{N}[fs, r], exp \in \text{Bias}[es, 2^{es-1}-1, r] \setminus \{-2^{es-1}+1, 2^{es-1}\})  \} \cup \{ \frac{x}{2^{fs}} \times 2^{exp}, f \in (\mathbb{N}[fs, r] \setminus \{0\}), exp = -2^{es-1}+2 \}  \} \cup \{0\}$ & $\text{IEEE754}[4,3,r]$ & \textcolor{red}{0}\textcolor{ForestGreen}{1001}\textcolor{orange}{010} & \textcolor{red}{1}\textcolor{ForestGreen}{1001}\textcolor{orange}{010} & \textcolor{red}{0}\textcolor{ForestGreen}{0111}\textcolor{orange}{100}\\
+            \cline{2-8}
+            %Morris
+            & \multirow{2}{*}{Morris} & IP & $\text{FloatingPoint}[\infty]$ & & \textcolor{red}{0}1\textcolor{blue}{.}\textcolor{orange}{010}/\textcolor{red}{0}\textcolor{ForestGreen}{10} & \textcolor{red}{1}1\textcolor{blue}{.}\textcolor{orange}{010}/\textcolor{red}{0}\textcolor{ForestGreen}{10} & \textcolor{red}{0}1\textcolor{blue}{.}\textcolor{orange}{100}/\textcolor{red}{0}\textcolor{ForestGreen}{0}\\
+            \cline{3-8}
+            & & VP/FP & $\{0\} \cup \bigcup_{G \in \mathbb{N}[g, r]} A_G=\{(-1)^{sign \times p} \times (1 + \frac{f}{2^{fs}}) \times 2^{exp}, es = min(G+1,n-g-1), t = max(G+1-(n-g-1), 0), e \in \mathbb{Z}[es+1,r], exp=e \times 2^t, fs=max(n-2-es-g,0), p=max(n-g-1-es-fs,0), sign \in \{0,1\}, f \in \mathbb{N}[fs, r] \}$ & $\text{Morris}[8,2,r]$ & \textcolor{blue}{01}\textcolor{red}{0}\textcolor{ForestGreen}{10}\textcolor{red}{0}\textcolor{orange}{01}&\textcolor{blue}{01}\textcolor{red}{0}\textcolor{ForestGreen}{10}\textcolor{red}{1}\textcolor{orange}{01}&\textcolor{blue}{00}\textcolor{red}{0}\textcolor{ForestGreen}{0}\textcolor{red}{0}\textcolor{orange}{010}\\
+            \cline{2-8}
+            %Posit
+            & \multirow{2}{*}{Posit} & IP & $\text{FloatingPoint}[\infty]$ & & \textcolor{red}{0}1\textcolor{blue}{.}\textcolor{orange}{010}/\textcolor{red}{0}\textcolor{ForestGreen}{10} & \textcolor{red}{1}1\textcolor{blue}{.}\textcolor{orange}{010}/\textcolor{red}{0}\textcolor{ForestGreen}{10} & \textcolor{red}{0}1\textcolor{blue}{.}\textcolor{orange}{100}/\textcolor{red}{0}\textcolor{ForestGreen}{0}\\
+            \cline{3-8}
+            & & VP/FP & $\{0\} \cup \bigcup_{kl \in \mathbb{N}, 0 < kl \leq n -2 } A_{kl}=\{(-1)^{sign} \times (1 + \frac{f}{2^{fs}}) \times 2^{exp+regime},t=max(0, es-(n-2-kl)), e \in \mathbb{N}[es-t, r], regime \in \{(kl-1) \times 2^{es},-kl \times 2 ^{es}\}, exp = e \times 2^t, fs = max(0, n-2-kl-es), f \in \mathbb{N}[fs, r], sign \in \{0,1\} \}$ & [8,2,r] & \textcolor{red}{0}\textcolor{blue}{10}\textcolor{ForestGreen}{10}\textcolor{orange}{010}&\textcolor{red}{1}0101110 & \textcolor{red}{0}\textcolor{blue}{10}\textcolor{ForestGreen}{00}\textcolor{orange}{100}\\
+            \hline
+        \end{tabular}%
+        }
+        %\end{center}
+        \label{tabel:nrssback}
+        \end{table}
+\end{frame}
+
+\begin{frame}
+    \frametitle{My proposals}
+    \begin{itemize}
+        \item Morris Hidden Exponent Bit \begin{equation}
+            \textcolor{red}{s_{f}}\textcolor{blue}{G\textsubscript{g-1}G\textsubscript{g-2}...G\textsubscript{0}}\textcolor{red}{s_{e}}\textcolor{brown}{e\textsubscript{es-1}e\textsubscript{es-2}...e\textsubscript{0}}\textcolor{orange}{f\textsubscript{fs-1}f\textsubscript{fs-2}...f\textsubscript{0}}
+        \end{equation}
+        \item Morris Hidden Exponent Bit with Bias G \begin{equation}
+            \textcolor{red}{s}\textcolor{blue}{G\textsubscript{g-1}G\textsubscript{g-2}...G\textsubscript{0}}\textcolor{brown}{e\textsubscript{es-1}e\textsubscript{es-2}...e\textsubscript{0}}\textcolor{orange}{f\textsubscript{fs-1}f\textsubscript{fs-2}...f\textsubscript{0}}
+        \end{equation}
+        \item Morris Hidden Exponent Bit with Unary G \begin{equation}
+            \textcolor{red}{s}\textcolor{blue}{r\textsubscript{0}r\textsubscript{1}...r\textsubscript{rs-2}\overline{r\textsubscript{rs-1}}}\textcolor{brown}{e\textsubscript{es-1}e\textsubscript{es-2}...e\textsubscript{0}}\textcolor{orange}{f\textsubscript{fs-1}f\textsubscript{fs-2}...f\textsubscript{0}}
+        \end{equation}
+    \end{itemize}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Morris Hidden Exponent Bit}
+The major problem of Morris floating numbers is the multiple ways of representing the same number.
+A solution for this is in borrowing the concept of hidden bit from mantissa.
+The $g$ field not only represents the $G$ value which dictates the exponent size but also the position of the most significant bit set in the exponent.
+If the value of the exponent size is kept as $G+1$, then the minimum absolute value of the exponent is 2 when $G=0$.
+\end{frame}
+\begin{frame}
+    \frametitle{Morris Hidden Exponent Bit}
+
+    $es=G-1$\\
+\begin{equation}
+    \text{exponent} = \begin{cases}
+    (-1)^{\text{exponent sign}} \times ( 2^{es} + \text{binaryExponent} ), & es \neq -1\\
+    0, & es=-1.
+\end{cases}
+\end{equation}
+\begin{equation}
+    \text{value} = \begin{cases}
+    0, & \text{all bits 0}\\
+    \text{NR}, & \text{first bit 1 and the rest 0s} \\
+    (-1)^{\text{sign}} \times 2^{\text{exponent}} \times (1 + \frac{\text{f}}{2^{\text{fs}}}), & \text{otherwise} \\
+  \end{cases}
+\end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Morris Hidden Exponent Bit with Bias G}
+One might argue that the problem of multiple representations is still not solved because even the
+exponent may have multiple values (for $es=-1$ the exponent sign does not matter).
+The problem stems from having a bit dedicated to the exponent sign.
+This is already solved in IEEE754 by using a bias value. A bias value $g$ is proposed.
+The exponent sign is the sign of $G$ and the exponent size is $es=|G|-1$.
+Another issue with Morris and MorrisHEB representations is that they do not have an order in binary form.
+A solution for this is to have the bits of the exponent negated when $G$ is negative.
+This makes it easy to implement a hardware compare unit.
+
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Morris Hidden Exponent Bit with Bias G}
+    $\textit{bias}=2^{g-1}-1$. 
+    $G=\textit{binary G} - \textit{bias}$
+    $es=|G|-1$
+    \begin{equation}
+        \text{exponent} = \begin{cases}
+        \text{signum(G)} \times ( 2^{es} + \text{binaryExponent} ), & es \neq -1\\
+        0, & es=-1.
+      \end{cases}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Morris Hidden Exponent Bit with Unary G}
+    \begin{equation}
+        \text{exponent size} = \begin{cases}
+        -k - 1, & k < 0 \\
+        k - 1, & k \ge 0.
+      \end{cases}
+    \end{equation}
+    \begin{equation}
+        k = \begin{cases}
+        -\text{NoC0}, & r_0=0 \\
+        \text{NoC1} - 1, & r_0=1.
+      \end{cases}
+    \end{equation}
+    \begin{equation}
+        \text{exponent} = \begin{cases}
+        \text{signum(k)} \times ( 2^{es} + \text{binaryExponent} ), & es \neq -1\\
+        0, & es=-1.
+      \end{cases}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Addition}
+
+\begin{figure}[tp]
+    \includegraphics[width=\textwidth]{media/add.jpg}
+\end{figure}
+
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Multiplication}
+    \begin{figure}[tp]
+    \includegraphics[width=\textwidth]{media/mul.jpg}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Exam Questiion}
+    Template: Given the hexadecimal string $0x\text{hex}$, what is the value of the number in the X NRS (number representation system)?
+    
+    Example: Given the binary string $0x5A$, what is the value of the number in the Morris Hidden Exponent Bit with Bias G (8, 2) representation?
+\end{frame}
diff --git a/slides/courses/2/text.tex b/slides/courses/2/text.tex
new file mode 100644
index 00000000..586f2322
--- /dev/null
+++ b/slides/courses/2/text.tex
@@ -0,0 +1,222 @@
+\subsection{Morse Code}
+
+\begin{frame}
+    \frametitle{Morse Code}
+        \begin{figure}
+            \caption{Morse Code By Rhey T. Snodgrass \& Victor F. Camp, 1922 - Own work based on: Intcode.png and International Morse Code.PNG, Public Domain, \url{https://commons.wikimedia.org/w/index.php?curid=3902977}}
+            \includegraphics[scale=0.32]{../media/International_Morse_Code.png}
+        \end{figure}
+    \note{
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Morse Code - Binary Storage}
+    \begin{itemize}
+        \item Primary:
+        \begin{itemize}
+            \item dots $1'b1$
+            \item space $1'b0$
+        \end{itemize}
+        \item Secondary:
+        \begin{itemize}
+            \item dash 3 dots $3'b111$
+            \item space inside letter between dash and dots $1'b0$
+            \item space between letters $3'b000$
+            \item space between words $7'b0000000$
+        \end{itemize}
+    \end{itemize}
+    \note{
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Morse Code - Binary Tree}
+    \begin{figure}
+        \caption{Morse Code Binary Tree}
+        \includegraphics[scale=0.45]{../media/Morse_code_tree3.png}
+    \end{figure}
+    \note{
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Morse Code - Why not?}
+    Give up on space:
+    \begin{itemize}
+        \item dots $1'b0$
+        \item dash $1'b1$
+    \end{itemize}
+    Used today in:
+    \begin{itemize}
+        \item transmission
+        \item storage
+    \end{itemize}
+\end{frame}
+
+\subsection{ASCII}
+
+\begin{frame}
+    \frametitle{ASCII}
+    \begin{itemize}
+        \item American Standard Code for Information Interchange
+        \item 7-bit encoding
+        \item 128 characters
+        \item 33 control characters
+        \item 95 printable characters
+    \end{itemize}
+    \note{
+    }
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{ASCII Table}
+
+    \begin{table}[h!]
+        \centering
+        \scalebox{0.8}{
+            \begin{tabular}{|c|c|c|c|c|c|c|c|}
+                \hline
+                \textbf{Hex} & \textbf{Char} & \textbf{Hex} & \textbf{Char} & \textbf{Hex} & \textbf{Char} & \textbf{Hex} & \textbf{Char} \\
+                \hline
+                0x00 & NUL & 0x10 & DLE & 0x20 & Space & 0x30 & 0 \\
+                0x01 & SOH & 0x11 & DC1 & 0x21 & ! & 0x31 & 1 \\
+                0x02 & STX & 0x12 & DC2 & 0x22 & " & 0x32 & 2 \\
+                0x03 & ETX & 0x13 & DC3 & 0x23 & \# & 0x33 & 3 \\
+                0x04 & EOT & 0x14 & DC4 & 0x24 & \$ & 0x34 & 4 \\
+                0x05 & ENQ & 0x15 & NAK & 0x25 & \% & 0x35 & 5 \\
+                0x06 & ACK & 0x16 & SYN & 0x26 & \& & 0x36 & 6 \\
+                0x07 & BEL & 0x17 & ETB & 0x27 & ' & 0x37 & 7 \\
+                0x08 & BS  & 0x18 & CAN & 0x28 & ( & 0x38 & 8 \\
+                0x09 & TAB & 0x19 & EM  & 0x29 & ) & 0x39 & 9 \\
+                0x0A & LF  & 0x1A & SUB & 0x2A & * & 0x3A & : \\
+                0x0B & VT  & 0x1B & ESC & 0x2B & + & 0x3B & ; \\
+                0x0C & FF  & 0x1C & FS  & 0x2C & , & 0x3C & < \\
+                0x0D & CR  & 0x1D & GS  & 0x2D & - & 0x3D & = \\
+                0x0E & SO  & 0x1E & RS  & 0x2E & . & 0x3E & > \\
+                0x0F & SI  & 0x1F & US  & 0x2F & / & 0x3F & ? \\
+                \hline
+            \end{tabular}
+        }
+        \caption{ASCII Table with Characters and Hexadecimal Values}
+        \label{tab:ascii_table}
+    \end{table}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{ASCII Table}
+
+    \begin{table}[h!]
+        \centering
+        \scalebox{0.8}{
+            \begin{tabular}{|c|c|c|c|c|c|c|c|}
+                \hline
+                \textbf{Hex} & \textbf{Char} & \textbf{Hex} & \textbf{Char} & \textbf{Hex} & \textbf{Char} & \textbf{Hex} & \textbf{Char} \\
+                \hline
+                0x40 & @  & 0x50 & P  & 0x60 & `  & 0x70 & p \\
+                0x41 & A  & 0x51 & Q  & 0x61 & a  & 0x71 & q \\
+                0x42 & B  & 0x52 & R  & 0x62 & b  & 0x72 & r \\
+                0x43 & C  & 0x53 & S  & 0x63 & c  & 0x73 & s \\
+                0x44 & D  & 0x54 & T  & 0x64 & d  & 0x74 & t \\
+                0x45 & E  & 0x55 & U  & 0x65 & e  & 0x75 & u \\
+                0x46 & F  & 0x56 & V  & 0x66 & f  & 0x76 & v \\
+                0x47 & G  & 0x57 & W  & 0x67 & g  & 0x77 & w \\
+                0x48 & H  & 0x58 & X  & 0x68 & h  & 0x78 & x \\
+                0x49 & I  & 0x59 & Y  & 0x69 & i  & 0x79 & y \\
+                0x4A & J  & 0x5A & Z  & 0x6A & j  & 0x7A & z \\
+                0x4B & K  & 0x5B & [  & 0x6B & k  & 0x7B & \{ \\
+                0x4C & L  & 0x5C & \textbackslash  & 0x6C & l  & 0x7C & | \\
+                0x4D & M  & 0x5D & ]  & 0x6D & m  & 0x7D & \} \\
+                0x4E & N  & 0x5E & \textasciicircum  & 0x6E & n  & 0x7E & \textasciitilde \\
+                0x4F & O  & 0x5F & \_  & 0x6F & o  & 0x7F & DEL \\
+                \hline
+            \end{tabular}
+        }
+        \caption{ASCII Table with Characters and Hexadecimal Values}
+        \label{tab:ascii_table}
+    \end{table}
+\end{frame}
+
+
+
+
+\subsection{Unicode}
+
+\begin{frame}
+    \frametitle{Unicode}
+    \begin{itemize}
+        \item 16-bit encoding
+        \item Contains ASCII
+    \end{itemize}
+    \note{
+    }
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Unicode Table}
+
+    \begin{table}[h!]
+        \centering
+        \scalebox{0.6}{
+        \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
+            \hline
+            \textbf{Hex} & \textbf{0} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} & \textbf{7} & \textbf{8} & \textbf{9} & \textbf{A} & \textbf{B} & \textbf{C} & \textbf{D} & \textbf{E} & \textbf{F} \\
+            \hline
+            0x0000 & NUL & SOH & STX & ETX & EOT & ENQ & ACK & BEL & BS  & TAB & LF  & VT  & FF  & CR  & SO  & SI  \\
+            0x0010 & DLE & DC1 & DC2 & DC3 & DC4 & NAK & SYN & ETB & CAN & EM  & SUB & ESC & FS  & GS  & RS  & US  \\
+            0x0020 & Space & ! & " & \# & \$ & \% & \& & ' & ( & ) & * & + & , & - & . & / \\
+            0x0030 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & : & ; & < & = & > & ? \\
+            0x0040 & @ & A & B & C & D & E & F & G & H & I & J & K & L & M & N & O \\
+            0x0050 & P & Q & R & S & T & U & V & W & X & Y & Z & [ & \textbackslash & ] & \textasciicircum & \_ \\
+            0x0060 & ` & a & b & c & d & e & f & g & h & i & j & k & l & m & n & o \\
+            0x0070 & p & q & r & s & t & u & v & w & x & y & z & \{ & | & \} & \textasciitilde & DEL \\
+            \hline
+        \end{tabular}
+        }
+        \caption{Unicode Table with Characters and Hexadecimal Values}
+        \label{tab:unicode_table}
+    \end{table}
+    \note{
+    }
+\end{frame}
+
+
+
+\begin{frame}
+    \frametitle{UTF-8}
+    \begin{itemize}
+        \item Unicode Transformation Format
+        \item 21-bit encoding
+        \item Variable-width encoding (1-4 bytes)
+    \end{itemize}
+    \note{
+    }
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{UTF-8 Encoding}
+    \begin{table}[h!]
+        \centering
+        \scalebox{0.8}{
+        \begin{tabular}{|c|c|c|}
+            \hline
+            \textbf{Bytes} & \textbf{Binary Format} & \textbf{Range} \\
+            \hline
+            1 byte  & 0xxxxxxx & U+0000 to U+007F \\
+            2 bytes & 110xxxxx 10xxxxxx & U+0080 to U+07FF \\
+            3 bytes & 1110xxxx 10xxxxxx 10xxxxxx & U+0800 to U+FFFF \\
+            4 bytes & 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx & U+10000 to U+10FFFF \\
+            \hline
+        \end{tabular}
+        }
+        \caption{UTF-8 Encoding Scheme}
+        \label{tab:utf8_encoding}
+    \end{table}
+
+    \note{
+    }
+\end{frame}
\ No newline at end of file
diff --git a/slides/courses/3/cache.tex b/slides/courses/3/cache.tex
new file mode 100644
index 00000000..a2a68c20
--- /dev/null
+++ b/slides/courses/3/cache.tex
@@ -0,0 +1,168 @@
+\begin{frame}
+    \frametitle{Cache Memory Propriety}
+    \begin{itemize}
+        \item Miss Rate
+        \item Average Memory Access Time
+        \item Hit Time
+        \item Miss Penalty
+        \item Cache Size
+        \item Block Size
+        \item Associativity
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Cache Memory Performence}
+    \begin{itemize}
+        \item Block Placement
+        \item Block Identification
+        \item Block Replacement
+        \item Write Strategy
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Block Placement}
+    \begin{itemize}
+        \item Cache memory contains a number $n$ of blocks.
+        \item RAM memory contains a number $m$ of blocks, ussualy $m \gg n$.
+    \end{itemize}
+    There are three main strategies for block placement:
+    \begin{itemize}
+        \item Direct Mapping (1-way set associative):\\
+        $p_{\text{cache}}=p_{\text{ram}} \mod n$
+        \item Fully Associative (n-way set associative):\\
+        $ \forall \ p_{\text{cache}}$
+        \item K-Set Associative (k-way set associative):\\
+        $\forall p_{\text{cache}} \ \in \text{set } (p_{\text{ram}} \mod \frac{n}{k})$
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Block Placement}
+    \note{
+    Draw the block placement for each strategy.
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Block Identification}
+    \begin{table}[h!]
+        \centering
+        \begin{tabular}{|p{5cm}|p{3cm}|p{3cm}|}
+            \hline
+            \multicolumn{3}{|p{11cm}|}{\textbf{Address}} \\
+            \hline
+            \multicolumn{2}{|p{8cm}|}{\textbf{Block Address}} & \multicolumn{1}{p{3cm}|}{\textbf{Block Offset}} \\
+            \hline
+            \multicolumn{1}{|p{5cm}|}{\textbf{Tag}} & \multicolumn{1}{|p{3cm}|}{\textbf{Index}} & \multicolumn{1}{p{3cm}|}{\textbf{Block Offset}} \\
+            \hline
+        \end{tabular}
+    \end{table}
+    \begin{itemize}
+        \item \textbf{Tag}: Identifies the block in memory.
+        \item \textbf{Index}: Identifies the set within the cache.
+        \item \textbf{Block Offset}: Specifies the exact byte within the block.
+        \item A valid bit is also included to indicate if the block is valid.
+        \item Increasing the associativity reduces the number of bits needed for the index but increases the bits required for the tag.
+        \item Increasing the number of sets raises the complexity of the hardware.
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Block Identification}
+    \note{
+        Illustrate the block identification process involving the TLB and the cache.
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Block Replacement}
+    \begin{itemize}
+        \item When a block needs to be replaced, the cache controller must determine which block to evict.
+        \item The replacement policy can be one of the following:
+        \begin{itemize}
+            \item Random
+            \item LRU (Least Recently Used) - this can be expensive to implement and is often only partially implemented
+            \item FIFO (First In, First Out)
+        \end{itemize}
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Block Replacement}
+    \note{
+        Explain the LRU (Least Recently Used) policy.
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Write Strategy}
+    \textbf{Write Strategy:}
+    \begin{itemize}
+        \item \textbf{Write Through:} Write to both the cache and RAM (L1 and L2).
+        \item \textbf{Write Back:} Write to the cache and mark the block as dirty. Write to RAM only when the block is replaced (L3).
+    \end{itemize}
+    
+    \textbf{Allocation Policy:}
+    \begin{itemize}
+        \item \textbf{Write Allocate:} Load the block into the cache and write to it (used with Write Back).
+        \item \textbf{No Write Allocate:} Write directly to RAM only (used with Write Through).
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Write Buffer}
+    \textbf{Write Buffer (Victim Buffer):}
+    \begin{itemize}
+        \item Stores write operations.
+        \item Has a fixed size (usually 8-16 entries). When full, the CPU must wait.
+        \item Helps avoid write stalls.
+        \item Must respond correctly when read-after-write operations access the same block.
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Write Strategy}
+    \note{
+        Explain the write buffer.
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Exam Question}
+    Template: Calculate the number of bits for the tag, index, and block offset for a cache memory with the following characteristics: X cache size, Y block placement, and Z block size.
+
+    Example: Calculate the number of bits for the tag, index, and block offset for a cache memory with the following characteristics: 64KB size, 4-way set associative, 64B block size.
+\end{frame}
+
+\begin{frame}
+    \frametitle{Types of Misses}
+    \begin{itemize}
+        \item Compulsory (cold start): The first access to a block.
+        \item Capacity: Occurs when the cache is too small.
+        \item Conflict: Happens when too many blocks map to the same set.
+        \item Coherence: Arises when another cache modifies the block.
+    \end{itemize}
+    \note{
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Types of Misses}
+    \note{
+        Explain the different types of cache misses.
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Cache Memory Benchmark}
+    \begin{itemize}
+        \item Average Memory Access Time (AMAT) = Hit Time + Miss Rate $\times$ Miss Penalty
+        \item Hit Time: The time taken to access the cache.
+        \item Miss Penalty: The time taken to access the main memory.
+        \item Miss Rate: The ratio of misses to the total number of accesses.
+        \item CPU Execution Time
+        \item Power Consumption
+    \end{itemize}
+\end{frame}
\ No newline at end of file
diff --git a/slides/courses/3/heirarchy.tex b/slides/courses/3/heirarchy.tex
new file mode 100644
index 00000000..d060942c
--- /dev/null
+++ b/slides/courses/3/heirarchy.tex
@@ -0,0 +1,64 @@
+\begin{frame}
+    \frametitle{Memory Hierarchy}
+    \begin{columns}
+        \column{0.5\textwidth}
+        \begin{itemize}
+            \item The CPU makes requests to memory for specific addresses.
+            \begin{itemize}
+                \item Hit: Data is found.
+                \item Miss: Data is not found.
+            \end{itemize}
+            \item For cache memory:
+            \begin{itemize}
+                \item Cache hit, cache miss.
+                \item Hardware manages cache levels (L1, L2, L3).
+                \item Uses blocks or lines as units of transfer.
+            \end{itemize}
+            \item For virtual memory:
+            \begin{itemize}
+                \item Page hit, page fault.
+                \item The operating system manages virtual memory.
+                \item Uses pages as units of transfer.
+            \end{itemize}
+        \end{itemize}
+
+        \column{0.5\textwidth}
+        \newsavebox{\asciimemheir}
+        \begin{lrbox}{\asciimemheir}
+            \begin{varwidth}{\maxdimen}
+            \VerbatimInput[fontsize=\scriptsize]{media/memheir.ascii}
+            \end{varwidth}
+        \end{lrbox}%
+
+        \begin{figure}[h]
+            \centering
+            \scalebox{0.7}{\usebox{\asciimemheir}}
+        \end{figure}
+
+    \end{columns}
+    \note{
+        The memory hierarchy creates the illusion of a large, fast memory system.
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Memory Locality}
+    The memory hierarchy creates the illusion of a large, fast memory system.
+    \begin{itemize}
+        \item \textbf{Temporal locality:} If a memory location is accessed, it is likely to be accessed again soon.
+        \item \textbf{Spatial locality:} If a memory location is accessed, nearby memory locations will likely be accessed soon after.
+    \end{itemize}
+    A full cache memory miss depends on two factors: the latency of the main memory (how long it takes to retrieve the first byte) and the bandwidth (how long it takes to retrieve the entire line or block).
+\end{frame}
+
+\begin{frame}
+    \frametitle{CPU Execution Time}
+    \begin{itemize}
+        \item \textbf{CPU execution time} = CPU clock cycles $\times$ Clock cycle time (CLKT)
+        \item \textbf{CPU clock cycles} = Instruction count (IC) $\times$ Cycles per instruction (CPI)
+        \item \textbf{With memory hierarchy:} CPU clock cycles = (CPU clock cycles $+$ Memory stall cycles) $\times$ Clock cycle time
+        \item \textbf{Memory stall cycles} = Number of misses $\times$ Miss penalty
+        \item \textbf{Memory stall cycles} = IC $\times$ Misses per instruction $\times$ Miss penalty
+        \item \textbf{Misses per instruction} = Miss rate $\times$ Memory accesses per instruction (this can be separated for read and write operations)
+    \end{itemize}
+\end{frame}
\ No newline at end of file
diff --git a/slides/courses/3/main.tex b/slides/courses/3/main.tex
new file mode 100644
index 00000000..dbf09c4c
--- /dev/null
+++ b/slides/courses/3/main.tex
@@ -0,0 +1,113 @@
+%\documentclass[notes,usenames,dvipsnames]{beamer}
+%\documentclass[notes]{beamer}       % print frame + notes
+%\documentclass[notes=only]{beamer}   % only notes
+\documentclass[usenames,dvipsnames]{beamer}             % only frames
+\usepackage{pgfpages}
+%\setbeameroption{show notes on second screen}
+%\setbeameroption{show notes}
+%subfigures
+\usepackage{caption}
+\usepackage{subcaption}
+
+%tables packages
+\usepackage{multirow}
+
+% math
+\usepackage{amsmath}
+
+% bash command
+\usepackage{graphicx}
+\usepackage{listings}
+% varbatim for ascii figures
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+%\usepackage{verbatim}
+\usepackage{lipsum} % for context
+\usepackage{fancyvrb}
+\usepackage{varwidth}
+
+
+\newsavebox{\asciigcn}
+
+% notes prefixed in pympress
+\addtobeamertemplate{note page}{}{\thispdfpagelabel{notes:\insertframenumber}}
+%theme used
+\usetheme{Madrid}
+%Information to be included in the title page:
+\title[Computer Architecture] %optional
+{Computer Architecture}
+
+\subtitle{Course no. 3 - Memory}
+
+\author[Ștefan-Dan Ciocîrlan] % (optional, for multiple authors)
+{}
+
+\institute[NUSTPB] % (optional)
+{
+  \inst{}%
+  National University of Science and Technology\\
+  POLITEHNICA Bucharest
+}
+
+\date[NUSTPB 2024] % (optional)
+{Computer Architecture}
+
+\logo{
+\includegraphics[height=0.9cm]{../../media/LOGO_UNSTPB_en.png}
+\includegraphics[height=0.9cm]{../../media/logoACSQ.jpeg}
+}
+
+
+% Roman numerals
+\newcommand*{\rom}[1]{\expandafter\@slowromancap\romannumeral #1@}
+%split
+\usepackage{amsmath}
+%colors
+%\usepackage[usenames,dvipsnames]{color} %loaded by the dcoument class
+%subfigure
+\usepackage{subcaption}
+% block over block uncover
+\setbeamercovered{invisible}
+%\setbeamercovered{transparent}
+
+%extra slide content
+\AtBeginSection[]
+{
+  \begin{frame}
+    \frametitle{Content}
+    \tableofcontents[currentsection]
+  \end{frame}
+}
+%notes or not
+%\setbeamertemplate{note page}[plain]
+
+\begin{document}
+
+\frame{\titlepage}
+
+\section{Heirarchy of Memory}
+\input{heirarchy.tex}
+
+\section{Cache Memory}
+\input{cache.tex}
+
+\section{Virtual Memory}
+\input{virtual.tex}
+
+%\section{Image Representation Systems}
+%\input{image.tex}
+
+%\section{Audio Representation Systems}
+%\input{audio.tex}
+
+\section{Q\&A}
+\begin{frame}
+\end{frame}
+
+%\begin{frame}
+%\frametitle{Table of Contents}
+%\tableofcontents
+%\end{frame}
+
+
+\end{document}
\ No newline at end of file
diff --git a/slides/courses/3/media/memheir.ascii b/slides/courses/3/media/memheir.ascii
new file mode 100644
index 00000000..fd0e73d1
--- /dev/null
+++ b/slides/courses/3/media/memheir.ascii
@@ -0,0 +1,19 @@
+            +---------------+
+            |      CPU      |
+            | +-----------+ |
+     +------+ | REGISTERS | |
+     |      | +-----------+ |
+     |      +---------------+
+     |
+     |
++----+----+     +---------+     +---------+
+| CACHE   |     | CACHE   |     | CACHE   |
+| LAYER 1 +-----+ LAYER 2 +-----+ LAYER 3 |
+|         |     |         |     |         |
++---------+     +---------+     +-----+---+
+                                      |
+         +---------+     +--------+   |
+         | STORAGE |     | MAIN   |   |
+         | HDD/SSD +-----+ MEMORY +---+
+         | FLASH   |     | RAM    |
+         +---------+     +--------+ 
\ No newline at end of file
diff --git a/slides/courses/3/virtual.tex b/slides/courses/3/virtual.tex
new file mode 100644
index 00000000..a733f211
--- /dev/null
+++ b/slides/courses/3/virtual.tex
@@ -0,0 +1,124 @@
+\begin{frame}
+    \frametitle{Virtual Memory}
+    Benefits of Virtual Memory:
+    \begin{itemize}
+        \item Efficient memory management
+        \item Enhanced protection
+        \item Shared memory capabilities
+        \item Process relocation
+        \item Faster process creation and startup time (without loading the entire process into memory)
+    \end{itemize}
+    
+    The process of obtaining the physical address from the virtual address is known as translation.
+    The operating system manages the virtual memory.
+\end{frame}
+
+\begin{frame}
+    \frametitle{Allocation Policies}
+    \begin{itemize}
+        \item\textbf{Paged Virtual Memory}: In this approach, the virtual memory is divided into fixed-size pages.
+        \item \textbf{Segmented Virtual Memory}: Here, the virtual memory is divided into segments of varying sizes
+        based on logical divisions.
+        \item \textbf{Combined Virtual Memory}: This method divides the virtual memory into segments, with each segment's
+        size being a multiple of the page size.
+    \end{itemize}
+    The miss penalty refers to the time it takes to retrieve a page from the disk. Therefore, the operating system
+    aims to minimize the number of misses by adopting a fully associative design, allowing pages to be placed
+    anywhere in the main memory.
+\end{frame}
+
+\begin{frame}
+    \frametitle{TLB vs MMU}
+
+    \begin{itemize}
+        \item \textbf{Translation Lookaside Buffer (TLB)}
+        \begin{itemize}
+            \item A specialized cache designed to speed up the translation from virtual to physical addresses.
+            \item Stores recent translations for quick access.
+            \item Checked first during the address translation process.
+            \item Very fast, but limited in size.
+        \end{itemize}
+        \item \textbf{Memory Management Unit (MMU)}
+        \begin{itemize}
+            \item A hardware component responsible for managing memory and caching operations.
+            \item Translates virtual addresses to physical addresses using page tables.
+            \item Manages page faults and enforces memory protection.
+            \item More complex and integrated directly into the CPU.
+        \end{itemize}
+    \end{itemize}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{TLB vs MMU}
+    \begin{table}[h!]
+        \centering
+        \begin{tabular}{|p{2cm}|p{4cm}|p{4cm}|}
+            \hline
+            \textbf{Component} & \textbf{Function} & \textbf{Characteristics} \\
+            \hline
+            TLB & Caches recent address translations & Fast, limited size \\
+            \hline
+            MMU & Manages memory and performs address translations & Complex, integrated into the CPU \\
+            \hline
+        \end{tabular}
+    \end{table}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Exam Questions}
+    \begin{itemize}
+        \item What happens to CPU execution time if we increase the block size to reduce the miss rate?
+        \item What happens to CPU execution time if we increase the cache size to reduce the miss rate?
+        \item What happens to CPU execution time if we increase the associativity to reduce the miss rate?
+        \item What happens to CPU execution time if we add multiple levels of cache?
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Block size increase}
+    We have a cache of size $n$ and a block size of $b_{0}$.
+    The miss rate is $m_{0}$ and the miss penalty is $p_{access}$ plus $p_{byte}$
+    for each byte in clock cycles. If we increase the block size to $b_{1}$,
+    the miss rate will be $m_{1}$. Compute Average memory access time (AMAT)
+    for both cases. The hit time is $h$.
+\end{frame}
+\begin{frame}
+    \frametitle{Block size increase}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Cache size increase}
+    We have a block size of $b_{0}$ and a miss rate of $m_{0}$.
+    The miss penalty is $p_{access}$ plus $p_{byte}$ for each byte in clock cycles.
+    If we increase the cache size to $n_{1}$, the miss rate will be $m_{1}$.
+    Compute Average memory access time (AMAT) for both cases.
+    The hit time is $h_{0}$ for the first cache and $h_{1}$ for the second one.
+\end{frame}
+\begin{frame}
+    \frametitle{Cache size increase}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Associativity increase}
+    We have a cache of size $n$ and a block size of $b_{0}$ with a $k_{0}$-way associativity.
+    The miss rate is $m_{0}$, and the miss penalty is $p_{access}$ plus $p_{byte}$ for
+    each byte in clock cycles. If we increase the associativity to $k_{1}$,
+    the miss rate will be $m_{1}$, and will increase the clock cycle time from $c_{0}$ to $c_{1}$.
+    Hit time is $h$ for both. Compute Average memory access time (AMAT) for both cases.
+\end{frame}
+\begin{frame}
+    \frametitle{Associativity increase}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Multiple levels of cache}
+    Suppose that in $m$ memory references, there are $m_{1}$ misses in the first cache and $m_{2}$ misses in the second cache.
+    The hit time for the first cache is $h_{1}$, and for the second cache is $h_{2}$.
+    The miss penalty for the second cache is $p_{2}$.
+    Compute the Average memory access time (AMAT).
+\end{frame}
+\begin{frame}
+    \frametitle{Multiple levels of cache}
+\end{frame}
\ No newline at end of file
diff --git a/slides/courses/4/bitwise.tex b/slides/courses/4/bitwise.tex
new file mode 100644
index 00000000..010ec912
--- /dev/null
+++ b/slides/courses/4/bitwise.tex
@@ -0,0 +1,13 @@
+\begin{frame}
+    \frametitle{Bitwise Operations}
+    \begin{itemize}
+        \item Bitwise operations: AND, OR, XOR, NOT, NEG
+        \item Shift operations: Left shift, Right shift, Arithmetic right shift
+        \item Rotate operations: Left rotate, Right rotate
+        \item Bit manipulation operations: Set, Clear, Toggle, Test
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Bitwise Operations}
+\end{frame}
\ No newline at end of file
diff --git a/slides/courses/4/comparison.tex b/slides/courses/4/comparison.tex
new file mode 100644
index 00000000..8b8401b5
--- /dev/null
+++ b/slides/courses/4/comparison.tex
@@ -0,0 +1,28 @@
+\begin{frame}
+    \frametitle{Comparison Operations}
+    \begin{itemize}
+        \item Less than (lt): $<$
+        \item Less than or equal (le): $\leq$
+        \item Greater than (gt): $>$
+        \item Greater than or equal (ge): $\geq$
+        \item Equal (eq): $==$
+        \item Not equal (ne): $\neq$
+    \end{itemize}
+    \note{
+    }
+\end{frame}
+
+\begin{frame}
+    \frametitle{Let's Do Some Math}
+    We only need to implement $lt$ and $eq$.
+    \begin{equation}
+        \begin{aligned}
+            & le = lt \lor eq \\
+            & gt = \neg le \\
+            & ge = \neg lt \\
+            & ne = \neg eq
+        \end{aligned}
+    \end{equation}
+    \note{
+    }
+\end{frame}
\ No newline at end of file
diff --git a/slides/courses/4/flags.tex b/slides/courses/4/flags.tex
new file mode 100644
index 00000000..24790254
--- /dev/null
+++ b/slides/courses/4/flags.tex
@@ -0,0 +1,40 @@
+\begin{frame}
+    \frametitle{Status Flags}
+    \begin{itemize}
+        \item Zero Flag (Z)
+        \item Carry Flag (C)
+        \item Sign Flag (S)
+        \item Overflow Flag (O)
+        \item Parity Flag (P)
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Mathematical Operations}
+    \begin{equation}
+        \begin{aligned}
+            &Z = \neg | \text{result} \\ 
+            &C = \text{result}[n+1] \\ 
+            &S = \text{result}[n] \\ 
+            &O = \text{Operation defined} \\ 
+            &P = \neg \oplus \text{result} 
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Overflow Flag}
+    \begin{itemize}
+        \item Adding two positive numbers results in a negative number.
+        \item Adding two negative numbers results in a positive number.
+        \item Multiplying two numbers produces a result that exceeds the available bits.
+        \item Subtracting a positive number from a negative number results in a positive number.
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Exam Question}
+    Template: What are the status flags for the following operation \( X \)?
+
+    Example: What are the status flags for the following operation (signed integers) \( 4'b0110 + 4'b1001 \)?
+\end{frame}
\ No newline at end of file
diff --git a/slides/courses/4/fpu.tex b/slides/courses/4/fpu.tex
new file mode 100644
index 00000000..213aea0a
--- /dev/null
+++ b/slides/courses/4/fpu.tex
@@ -0,0 +1,5 @@
+\begin{frame}
+    \frametitle{Floating Point Unit}
+    \note{
+    }
+\end{frame}
diff --git a/slides/courses/4/intro.tex b/slides/courses/4/intro.tex
new file mode 100644
index 00000000..2e9649f9
--- /dev/null
+++ b/slides/courses/4/intro.tex
@@ -0,0 +1,28 @@
+\begin{frame}
+    \frametitle{ALU Structure}
+    \begin{columns}
+        \column{0.5\textwidth}
+        \begin{itemize}
+            \item ALU (Arithmetic Logic Unit) is a combinational circuit.
+            \item It performs arithmetic and logic operations.
+            \item There is a control input for selecting the desired operation.
+            \item It might include status flags.
+            \item The ALU can vary in cycle time and complexity.
+        \end{itemize}
+
+        \column{0.5\textwidth}
+            \newsavebox{\asciialu}
+            \begin{lrbox}{\asciialu}
+                \begin{varwidth}{\maxdimen}
+                \VerbatimInput[fontsize=\scriptsize]{media/alu.ascii}
+                \end{varwidth}
+            \end{lrbox}%
+
+            \begin{figure}[h]
+                \centering
+                \scalebox{0.7}{\usebox{\asciialu}}
+            \end{figure}
+    \end{columns}
+    \note{
+    }
+\end{frame}
diff --git a/slides/courses/4/main.tex b/slides/courses/4/main.tex
new file mode 100644
index 00000000..ecddae01
--- /dev/null
+++ b/slides/courses/4/main.tex
@@ -0,0 +1,116 @@
+%\documentclass[notes,usenames,dvipsnames]{beamer}
+%\documentclass[notes]{beamer}       % print frame + notes
+%\documentclass[notes=only]{beamer}   % only notes
+\documentclass[usenames,dvipsnames]{beamer}             % only frames
+\usepackage{pgfpages}
+%\setbeameroption{show notes on second screen}
+%\setbeameroption{show notes}
+%subfigures
+\usepackage{caption}
+\usepackage{subcaption}
+
+%tables packages
+\usepackage{multirow}
+
+% math
+\usepackage{amsmath}
+
+% bash command
+\usepackage{graphicx}
+\usepackage{listings}
+% varbatim for ascii figures
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+%\usepackage{verbatim}
+\usepackage{lipsum} % for context
+\usepackage{fancyvrb}
+\usepackage{varwidth}
+
+
+\newsavebox{\asciigcn}
+
+% notes prefixed in pympress
+\addtobeamertemplate{note page}{}{\thispdfpagelabel{notes:\insertframenumber}}
+%theme used
+\usetheme{Madrid}
+%Information to be included in the title page:
+\title[Computer Architecture] %optional
+{Computer Architecture}
+
+\subtitle{Course no. 4 - Arithmetic Logic Unit (ALU)}
+
+\author[Ștefan-Dan Ciocîrlan] % (optional, for multiple authors)
+{}
+
+\institute[NUSTPB] % (optional)
+{
+  \inst{}%
+  National University of Science and Technology\\
+  POLITEHNICA Bucharest
+}
+
+\date[NUSTPB 2024] % (optional)
+{Computer Architecture}
+
+\logo{
+\includegraphics[height=0.9cm]{../../media/LOGO_UNSTPB_en.png}
+\includegraphics[height=0.9cm]{../../media/logoACSQ.jpeg}
+}
+
+
+% Roman numerals
+\newcommand*{\rom}[1]{\expandafter\@slowromancap\romannumeral #1@}
+%split
+\usepackage{amsmath}
+%colors
+%\usepackage[usenames,dvipsnames]{color} %loaded by the dcoument class
+%subfigure
+\usepackage{subcaption}
+% block over block uncover
+\setbeamercovered{invisible}
+%\setbeamercovered{transparent}
+
+%extra slide content
+\AtBeginSection[]
+{
+  \begin{frame}
+    \frametitle{Content}
+    \tableofcontents[currentsection]
+  \end{frame}
+}
+%notes or not
+%\setbeamertemplate{note page}[plain]
+
+\begin{document}
+
+\frame{\titlepage}
+
+\section{ALU structure}
+\input{intro.tex}
+
+\section{Bitwise Operations} 
+\input{bitwise.tex}
+
+\section{Arithmetic Operations}
+\input{number.tex}
+
+\section{Comparison Operations}
+\input{comparison.tex}
+
+\section{Status Flags}
+\input{flags.tex}
+
+% \section{Floating Point Unit (FPU)}
+% \input{fpu.tex}
+
+\section{Q\&A}
+\begin{frame}
+\end{frame}
+
+%\begin{frame}
+%\frametitle{Table of Contents}
+%\tableofcontents
+%\end{frame}
+
+
+\end{document}
\ No newline at end of file
diff --git a/slides/courses/4/media/4-bit_carry_lookahead_adder.png b/slides/courses/4/media/4-bit_carry_lookahead_adder.png
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diff --git a/slides/courses/4/media/Kogge-stone-8-bit.png b/slides/courses/4/media/Kogge-stone-8-bit.png
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diff --git a/slides/courses/4/media/adders_comparison.jpg b/slides/courses/4/media/adders_comparison.jpg
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diff --git a/slides/courses/4/media/alu.ascii b/slides/courses/4/media/alu.ascii
new file mode 100644
index 00000000..df4c880a
--- /dev/null
+++ b/slides/courses/4/media/alu.ascii
@@ -0,0 +1,23 @@
++-----------+     +-----------+
+|           |     |           |
+| OPERAND 1 |     | OPERAND 2 |
+|           |     |           |
++-----+-----+     +-----+-----+
+      |                 |
+      |                 |
+      v                 v 
++-----------------------------+
+|            ALU              |     +-----------+
+|                             |     |           |
+|                             |<----+ OPERATION |
+|                             |     | SELECTOR  |
+|                             |     +-----------+ 
++--------------+--------------+ 
+    |                   | 
+    |                   |
+    v                   v 
++--------+          +--------+
+|        |          |        |
+| RESULT |          | FLAGS  |
+|        |          |        |
++--------+          +--------+
\ No newline at end of file
diff --git a/slides/courses/4/media/cla_tree.jpg b/slides/courses/4/media/cla_tree.jpg
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diff --git a/slides/courses/4/media/full-adder-gates.png b/slides/courses/4/media/full-adder-gates.png
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diff --git a/slides/courses/4/media/ripple-carry.png b/slides/courses/4/media/ripple-carry.png
new file mode 100644
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diff --git a/slides/courses/4/media/ripple-carry4bit.png b/slides/courses/4/media/ripple-carry4bit.png
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diff --git a/slides/courses/4/number.tex b/slides/courses/4/number.tex
new file mode 100644
index 00000000..714cf02e
--- /dev/null
+++ b/slides/courses/4/number.tex
@@ -0,0 +1,337 @@
+\subsection{ADD/SUB}
+\begin{frame}
+    \frametitle{Addition/Subtraction}
+    A selector can be designed so that either addition (ADD) or subtraction (SUB)
+    can be performed using the same hardware when working with two's complement representations.
+    Let's define the operands $x$ and $y$ with the following notations:
+    \begin{equation}
+        \begin{aligned}
+            &\text{result} =
+            \begin{cases}
+                x + y, & \text{if ADD} \\
+                x + (-y), & \text{if SUB}
+            \end{cases}
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Addition/Subtraction}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Adder}
+    The various types of adder implementations in hardware include:
+    \begin{itemize}
+        \item Ripple Carry Adder
+        \item Carry Lookahead Adder
+            \begin{itemize}
+                \item Carry Skip Adder (bypass)
+                \item Carry Select Adder (multiplexer)
+            \end{itemize}
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Half adder}
+ A half adder is a combinational circuit that adds two bits and produces a sum and carry bits.
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.7\textwidth]{media/half-adder-gates.png}
+        \caption{Half Adder}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Full adder}
+ A full adder is a combinational circuit that adds three bits and produces a sum and carry bits.
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.7\textwidth]{media/full-adder-gates.png}
+        \caption{Full Adder}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Ripple Carry Adder}
+ The simplest adder is the ripple carry adder.
+ It is composed of a chain of full adders.
+ The carry-out of one full adder is the carry-in of the next full adder.
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.7\textwidth]{media/ripple-carry.png}
+        \caption{Ripple Carry Adder}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Let's do some math}
+    \begin{equation}
+        \begin{aligned}
+            &sum[0]=a[0] \oplus b[0] \oplus c_{in}[0]\\
+            &c_{in}[1]=c_{out}[0]=(a[0]\ \& \ b[0]) \ | \ (c_{in}[0] \ \& \ (a[0] \oplus b[0]))\\
+            &sum[i]=a[i] \oplus b[i] \oplus c_{in}[i]\\
+            &c_{in}[i+1]=c_{out}[i]=(a[i] \ \& \ b[i]) \ | \ (c_{in}[i] \ \& \ (a[i] \oplus b[i]))\\
+            &G[i]=a[i] \ \& \ b[i]\\
+            &P[i]=a[i] \oplus b[i]\\
+            &c_{in}[i+1]=c_{out}[i]=G[i] \ | \ (c_{in}[i] \ \& \ P[i])
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Let's do some math}
+    \begin{equation}
+        \begin{aligned}
+            & \text{Ripple Carry Adder number of gates for carry out: } 3 \times n\\
+            & \text{Carry Lookahead Adder number of gates for carry out: } 2 \times n\\
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+
+\begin{frame}
+    \frametitle{Carry Lookahead Adder}
+ The carry lookahead adder is a more complex adder that reduces the time to calculate the carry-out of each full adder.
+ It is composed of two main blocks:
+    \begin{itemize}
+        \item Generate block
+        \item Propagate block
+    \end{itemize}
+ The following formula calculates the carry-out and sum of each full adder:
+    \begin{equation}
+        \begin{aligned}
+            &c_{out}[i]=G[i] \ | \ (P[i] \ \& \ c_{in}[i])\\
+            &sum[i]=P[i] \oplus c_{in}[i]
+        \end{aligned}
+    \end{equation}
+
+\end{frame}
+
+\begin{frame}
+    \frametitle{Carry Lookahead Adder}
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.5\textwidth, angle=270 ]{media/adders_comparison.jpg}
+        \caption{Carry Lookahead Adder vs Ripple Carry Adder}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Let's do some math}
+    \begin{equation}
+ \resizebox{0.9\textwidth}{!}{%
+            $
+            \begin{aligned}
+                &c_{in}[i+1]=c_{out}[i]=G[i] \ | \  (P[i] \ \& \ c_{in}[i])\\
+                &c_{in}[i+1]=G[i] \ | \  (P[i] \ \& \ (G[i-1]  \ | \  (P[i-1] \ \& \ c_{in}[i-1])))\\
+                &c_{in}[i+1]=G[i] \ | \  (P[i] \ \& \ (G[i-1]  \ | \  (P[i-1] \ \& \ (G[i-2]  \ | \  P[i-2] \ \& \ c_{in}[i-2]))))\\
+                &c_{in}[i+1]=G[i] \ | \  (P[i] \ \& \ G[i-1])  \ | \  (P[i] \ \& \ P[i-1] \ \& \ (G[i-2]  \ | \  (P[i-2] \ \& \ c_{in}[i-2])))\\
+                &c_{in}[i+1]=G[i] \ | \  (\sum_{j=1}^{i} (\prod_{k=j}^{i} P[k]) \  \& \ G[j-1]) \ | \  ((\prod_{k=0}^{i} P[k]) \ \& \ c_{in}[0])\\
+                &\text{Notation: } i \geq j \text{ , } P[i, j] = \prod_{k=j}^{i} P[k] \\
+                &c_{in}[i+1]=G[i] \ | \  (\sum_{j=1}^{i} P[i, j] \  \& \ G[j-1]) \ | \  (P[i, 0] \ \& \ c_{in}[0])\\
+            \end{aligned}$%
+ }
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Carry Lookahead Adder Tree}
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.5\textwidth]{media/cla_tree.jpg}
+        \caption{Carry Lookahead Adder Tree}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Carry Lookahead Adder Kogge-Stone}
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.6\textwidth]{media/Kogge-stone-8-bit.png}
+        \caption{CLA Kogge-Stone}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Carry Lookahead Adder Brent-Kung}
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.6\textwidth]{media/Brent-kung-8-bit.png}
+        \caption{CLA Brent-Kung}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Carry Skip Adder (bypass)}
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.9\textwidth]{media/BCSAdder16Bit.png}
+        \caption{CSA 16-bit By Tibor89 - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=31748842}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Carry Select Adder (multiplexer)}
+    \begin{figure}
+        \centering
+        \includegraphics[width=0.9\textwidth]{media/Carry-select-adder-fixed-size.png}
+        \caption{CSelectA 16-bit}
+    \end{figure}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Exam Questions}
+ Template: Given the x-bit adder of type y, calculate the number of gates for the carry-out.
+
+ Example: Given the 8-bit ripple carry adder, calculate the number of gates for the carry-out.
+
+ Example: Given the 16-bit carry-lookahead adder Brent-Kung on 4-bits with Carry Skip adder, calculate the number of gates for the carry-out.
+\end{frame}
+
+\subsection{MUL}
+
+\begin{frame}
+    \frametitle{Multiplication}
+    \begin{equation}
+        \begin{aligned}
+            &P = M \times R\\
+            &P=\sum_{i=0}^{n-1} (M \times R[i] \times 2^i)\\
+            &P=\sum_{i=0}^{n-1} ((M \ \& \ \{n \times R[i]\}) \ll i)\\
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Let's do some math}
+    \begin{equation}
+        \begin{aligned}
+            &P = M \times R\\
+            &P = M \times 8'b0111_0011\\
+            &P = M \times (2^6 + 2^5 + 2^4 + 2^1 + 2^0)\\
+            &P = M \times (2^7 - 2^5 + 2^2 - 2^0)\\
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Booth's Algorithm}
+ Booth's algorithm is a multiplication algorithm that multiplies two signed binary numbers in two's complement notation.
+ The algorithm reduces the number of additions required for the multiplication of two numbers.
+    \begin{equation}
+        \begin{aligned}
+            &|M|=|R|=n\\
+            &|P|=2 \times n + 1\\
+            &|M_{+}|=|M_{-}|=2 \times n+1\\
+            &M_{+}=M \ll (n + 1)\\
+            &M_{-}=-M \ll (n + 1)\\
+            &-M=\overline{M}+1\\
+            &|-M|=n\\
+            &M_{+}=\{M, 0\}\\
+            &M_{-}=\{-M, 0\}\\
+            &P=\{0,R,1'b0\}\\
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Booth's Algorithm}
+ Booth's Algorithm steps:
+    \begin{equation}
+        \begin{aligned}
+            &P=\{0,R,1'b0\}\\
+            &\text{for n steps}\\
+            &\begin{cases}
+ \text{if } P[1:0]=2'b01 \text{ then } P=P+M_{+}\\
+ \text{if } P[1:0]=2'b10 \text{ then } P=P+M_{-}\\
+ \text{otherwise } P=P\\
+            \end{cases}\\
+            &P=P \ggg 1\\
+            &\text{end for}\\
+            &result=P[(2 \times n):1]
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Exam Questions}
+ Template: Given the x-bit Booth's algorithm, how many additions are for $M \times R$?
+
+ Example: Given the 8-bit Booth's algorithm, how many additions are for $8'b0010_0110 \times 8'b1111_0110$?
+\end{frame}
+
+\subsection{DIV/MOD}
+
+\begin{frame}
+    \frametitle{DIV/MOD Debate}In computer engineering, integer division is a debated issue.
+    The next formulas define integer division:
+    \begin{equation}
+        \begin{aligned}
+ \text{Dividend} &= \text{Divisor} \times \text{Quotient} + \text{Remainder}\\
+ \text{Remainder} &= \text{Dividend} - \text{Divisor} \times \text{Quotient} 
+        \end{aligned}
+    \end{equation}
+The following operations might help $(-5)/2$ and $5/(-2)$ might help to understand the debate.
+ Taking the first one, $5=2 \times -3 + 1$, but another solution is $5=2 \times -2 + (-1)$.
+ A similar choice is for the second one, $-5=-2 \times 3 + 1$ and $-5=-2 \times 2 + (-1)$.
+\end{frame}
+
+\begin{frame}
+    \frametitle{DIV/MOD Debate}
+We can categorize the following types of divisions \footnote{Modulo operation. (2022, November 16). In Wikipedia. https://en.wikipedia.org/wiki/Modulo\_operation}:
+    \begin{itemize}
+        \item Truncated division. $\text{Quotient}=[\frac{\text{Dividend}}{\text{Divisor}}]$ (round down). The remainder and the dividend have equal signs.
+        \item Floored division. $\text{Quotient}=\lfloor \frac{\text{Dividend}}{\text{Divisor}} \rfloor$ (integer part). The remainder and the divisor have equal signs.
+        \item Euclidean division.  The remainder is always positive. \par  $\text{Quotient}=\text{sign}(\text{Divisor}) \times \lfloor \frac{\text{Dividend}}{|\text{Divisor}|} \rfloor = \begin{cases}
+             \lfloor \frac{\text{Dividend}}{\text{Divisor}} \rfloor & \text{if } \text{Divisor}>0\\
+             \lceil \frac{\text{Dividend}}{\text{Divisor}} \rceil & \text{if } \text{Divisor}<0
+        \end{cases}$
+        \item Round division. $\text{Quotient}=\text{round}(\frac{\text{Dividend}}{\text{Divisor}})$
+        \item Ceiling division. $\text{Quotient}=\lceil \frac{\text{Dividend}}{\text{Divisor}} \rceil$. The remainder and the divisor have opposite signs.
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Let's do some math for unsigned numbers}
+    \begin{equation}
+        \begin{aligned}
+            &Q = \text{Dividend}\\
+            &M = \text{Divisor}\\
+            &R = \text{Remainder, Quotient}\\
+            &|Q|=|M|=n\\
+            &|R|=2 \times n\\
+            &R=\{0,Q\}\\
+            &|M_{+}|=|M_{-}|=2 \times n\\
+            &M_{+}=M \ll n\\
+            &M_{-}=-M \ll n\\
+            &-M=\overline{M}+1\\
+            &|-M|=n\\
+            &M_{+}=\{M, 0\}\\
+            &M_{-}=\{-M, 0\}\\
+        \end{aligned}
+    \end{equation}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Booth's Algorithm for non-restoring division for unsigned numbers}
+    \begin{equation}
+        \begin{aligned}
+            &R=\{0,Q\}\\
+            &\text{for n steps}\\
+            &R=R \ll 1\\
+            &\text{if } R[2 \times n - 1]=\begin{cases}
+                1'b1 \text{ then } R=R+M_{+}\\
+                1'b0 \text{ then } R=R+M_{-}\\
+            \end{cases}\\
+            &\text{if } R[2 \times n - 1]=\begin{cases}
+                1'b1 \text{ then } R[0]=0\\
+                1'b0 \text{ then } R[0]=1\\
+            \end{cases}\\
+            &\text{end for}\\
+            &\text{if } R[2 \times n - 1]=1'b1 \text{ then } R=R+M_{+}\\
+            &\text{Quotient}=R[n-1:0]\\
+            &\text{Remainder}=R[2 \times n -1 : n]
+        \end{aligned}
+    \end{equation}
+\end{frame}
\ No newline at end of file