From 0031937a1991087c7c659803c51d7f392925c792 Mon Sep 17 00:00:00 2001
From: NotXia <35894453+NotXia@users.noreply.github.com>
Date: Fri, 22 Sep 2023 16:42:08 +0200
Subject: [PATCH] Add SMM finite numbers

---
 .gitignore                                    |   8 +
 .../img/_inherent_error.pdf                   | Bin 0 -> 29740 bytes
 .../img/floatingpoint_range.png               | Bin 0 -> 10045 bytes
 .../img/inherent_error.drawio                 |  88 ++++++++
 .../img/machine_eps.png                       | Bin 0 -> 5362 bytes
 .../main.tex                                  |  51 +++++
 .../sections/finite_numbers.tex               | 201 ++++++++++++++++++
 7 files changed, 348 insertions(+)
 create mode 100644 .gitignore
 create mode 100644 statistical-and-mathematical-methods-for-ai/img/_inherent_error.pdf
 create mode 100644 statistical-and-mathematical-methods-for-ai/img/floatingpoint_range.png
 create mode 100644 statistical-and-mathematical-methods-for-ai/img/inherent_error.drawio
 create mode 100644 statistical-and-mathematical-methods-for-ai/img/machine_eps.png
 create mode 100644 statistical-and-mathematical-methods-for-ai/main.tex
 create mode 100644 statistical-and-mathematical-methods-for-ai/sections/finite_numbers.tex

diff --git a/.gitignore b/.gitignore
new file mode 100644
index 0000000..ed96c40
--- /dev/null
+++ b/.gitignore
@@ -0,0 +1,8 @@
+*.synctex.gz
+*.log
+*.fls
+*.fdb_latexmk
+*.aux
+*.toc
+*.out
+[!_]*.pdf
\ No newline at end of file
diff --git a/statistical-and-mathematical-methods-for-ai/img/_inherent_error.pdf b/statistical-and-mathematical-methods-for-ai/img/_inherent_error.pdf
new file mode 100644
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diff --git a/statistical-and-mathematical-methods-for-ai/main.tex b/statistical-and-mathematical-methods-for-ai/main.tex
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+\documentclass[11pt]{article}
+\usepackage[margin=3cm]{geometry}
+\usepackage{graphicx}
+\usepackage{amsmath, amsfonts, amssymb, amsthm, mathtools}
+\usepackage{hyperref}
+\usepackage[nameinlink]{cleveref}
+\usepackage[all]{hypcap} % Links hyperref to object top and not caption
+\usepackage[inline]{enumitem}
+\usepackage{marginnote}
+
+\title{Statistical and Mathematical Methods for Artificial Intelligence}
+\date{2023 -- 2024}
+
+\hypersetup{
+    colorlinks,
+    citecolor=black,
+    filecolor=black,
+    linkcolor=black,
+    urlcolor=black,
+    linktoc=all
+}
+
+\setlist[description]{labelindent=\parindent}   % Indents `description`
+
+\newtheorem{example}{Example}[section]
+
+
+\begin{document}
+
+    \makeatletter
+    \begin{titlepage}
+        \centering
+        \vspace*{\fill}
+        \huge
+        \textbf{\@title}
+        \vspace*{\fill}
+
+        \Large
+        Academic Year \@date\\
+        Alma Mater Studiorum $\cdot$ University of Bologna
+        \vspace*{1cm}
+    \end{titlepage}
+    \makeatother
+    \pagenumbering{roman}
+    \tableofcontents
+    \newpage
+    \pagenumbering{arabic}
+
+    \input{sections/finite_numbers.tex}
+
+\end{document}
\ No newline at end of file
diff --git a/statistical-and-mathematical-methods-for-ai/sections/finite_numbers.tex b/statistical-and-mathematical-methods-for-ai/sections/finite_numbers.tex
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+++ b/statistical-and-mathematical-methods-for-ai/sections/finite_numbers.tex
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+\section{Finite numbers}
+
+
+
+\subsection{Sources of error}
+
+\begin{description}
+    \item[Measure error] 
+        Precision of the measurement instrument.
+
+    \item[Arithmetic error] 
+        Propagation of rounding errors in each step of an algorithm.
+
+    \item[Truncation error] 
+        Approximating an infinite procedure into a finite number of iterations.
+
+    \item[Inherent error] 
+        Caused by the finite representation of the data (floating-point).
+        \begin{figure}[h]
+            \centering
+            \includegraphics[width=0.6\textwidth]{img/_inherent_error.pdf}
+            \caption{Inherent error visualization}
+        \end{figure}
+\end{description}
+
+
+
+\subsection{Error measurement}
+
+Let $x$ be a value and $\hat{x}$ its approximation. Then:
+\begin{description}
+    \item[Absolute error] 
+        \begin{equation}
+            E_{a} = \hat{x} - x
+        \end{equation}
+        Note that, out of context, the absolute error is meaningless.
+    \item[Relative error] 
+        \begin{equation}
+            E_{a} = \frac{\hat{x} - x}{x}
+        \end{equation}
+\end{description}
+
+
+
+\subsection{Representation in base \texorpdfstring{$\beta$}{B}}
+
+Let $\beta \in \mathbb{N}_{> 1}$ be the base.
+Each $x \in \mathbb{R} \smallsetminus \{0\}$ can be uniquely represented as:
+\begin{equation} \label{eq:finnum_b_representation}
+    x = \texttt{sign}(x) \cdot (d_1\beta^{-1} + d_2\beta^{-2} + \dots d_n\beta^{-n})\beta^p
+\end{equation}
+where:
+\begin{itemize}
+    \item $0 \leq d_i \leq \beta-1$
+    \item $d_1 \neq 0$
+    \item starting from an index $i$, not all $d_j$ ($j \geq i$) are equal to $\beta-1$
+\end{itemize}
+%
+\Cref{eq:finnum_b_representation} can be represented using the normalized scientific notation as:
+\begin{equation}
+    x = \pm (0.d_1d_2\dots) \beta^p
+\end{equation}
+where $0.d_1d_2\dots$ is the \textbf{mantissa} and $\beta^p$ the \textbf{exponent}.
+
+
+
+\subsection{Floating-point}
+
+A floating-point system $\mathcal{F}(\beta, t, L, U)$ is defined by the parameters:
+\begin{itemize}
+    \item $\beta$: base
+    \item $t$: precision (number of digits in the mantissa)
+    \item $[L, U]$: range of the exponent
+\end{itemize}
+%
+Each $x \in \mathcal{F}(\beta, t, L, U)$ can be represented in its normalized form:
+\begin{eqnarray}
+    x = \pm (0.d_1d_2 \dots d_t) \beta^p & L \leq p \leq U
+\end{eqnarray}
+\begin{example}
+    In $\mathcal{F}(10, 5, -3, 3)$, $x=12.\bar{3}$ is represented as:
+    \begin{equation*}
+        \texttt{fl}(x) = + 0.12333 \cdot 10^2
+    \end{equation*}
+\end{example}
+
+
+\subsubsection{Numbers distribution}
+Given a floating-point system $\mathcal{F}(\beta, t, L, U)$, the total amount of representable numbers is:
+\begin{equation*}
+    2(\beta-1) \beta^{t-1} (U-L+1)+1
+\end{equation*}
+%
+Representable numbers are more sparse towards the exponent upper bound and more dense towards the lower bound.
+It must be noted that there is an underflow area around 0.
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.8\textwidth]{img/floatingpoint_range.png}
+    \caption{Floating-point numbers in $\mathcal{F}(2, 3, -1, 2)$}
+\end{figure}
+
+
+\subsubsection{Numbers representation}
+Given a floating-point system $\mathcal{F}(\beta, t, L, U)$, the representation of $x \in \mathbb{R}$ can result in:
+\begin{description}
+    \item[Exact representation] 
+        if $p \in [L, U]$ and $d_i=0$ for $i>t$.
+
+    \item[Approximation] 
+        if $p \in [L, U]$ but $d_i$ may not be 0 for $i>t$. 
+        In this case, the representation is obtained by truncating or rounding the value.
+
+    \item[Underflow] 
+        if $p < L$. In this case, the values is approximated as 0.
+
+    \item[Overflow] 
+        if $p > U$. In this case, an exception is usually raised.
+\end{description}
+
+
+\subsubsection{Machine precision}
+Machine precision $\varepsilon_{\text{mach}}$ determines the accuracy of a floating-point system.
+Depending on the approximation approach, machine precision can be computes as:
+\begin{description}
+    \item[Truncation] $\varepsilon_{\text{mach}} = \beta^{1-t}$
+    \item[Rounding] $\varepsilon_{\text{mach}} = \frac{1}{2}\beta^{1-t}$
+\end{description}
+Therefore, rounding results in more accurate representations.
+
+$\varepsilon_{\text{mach}}$ is the smallest distance among the representable numbers (\Cref{fig:finnum_eps}).
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.2\textwidth]{img/machine_eps.png}
+    \caption{Visualization of $\varepsilon_{\text{mach}}$ in $\mathcal{F}(2, 3, -1, 2)$}
+    \label{fig:finnum_eps}
+\end{figure}\\
+%
+In alternative, $\varepsilon_{\text{mach}}$ can be defined as the smallest representable number such that:
+\begin{equation*}
+    \texttt{fl}(1 + \varepsilon_{\text{mach}}) > 1.
+\end{equation*}
+
+
+\subsubsection{IEEE standard}
+IEEE 754 defines two floating-point formats:
+\begin{description}
+    \item[Single precision] Stored in 32 bits. Represents the system $\mathcal{F}(2, 24, -128, 127)$.
+        \begin{center}
+            \small
+            \begin{tabular}{|c|c|c|}
+                \hline
+                1 (sign) & 8 (exponent) & 23 (mantissa) \\
+                \hline
+            \end{tabular}
+        \end{center}
+
+    \item[Double precision] Stored in 64 bits. Represents the system $\mathcal{F}(2, 53, -1024, 1023)$.
+        \begin{center}
+            \small
+            \begin{tabular}{|c|c|c|}
+                \hline
+                1 (sign) & 11 (exponent) & 52 (mantissa) \\
+                \hline
+            \end{tabular}
+        \end{center}
+\end{description}
+As the first digit of the mantissa is always 1, it does not need to be stored.
+Moreover, special configurations are reserved to represent \texttt{Inf} and \texttt{NaN}.
+
+
+\subsubsection{Floating-point arithmetic}
+Let:
+\begin{itemize}
+    \item $+: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be a real numbers operation.
+    \item $\oplus: \mathcal{F} \times \mathcal{F} \rightarrow \mathcal{F}$ be the corresponding operation in a floating-point system.
+\end{itemize}
+%
+To compute $x \oplus y$, a machine:
+\begin{enumerate}
+    \item Calculates $x + y$ in a high precision register (still approximated, but more precise than the storing system)
+    \item Stores the result as $\texttt{fl}(x + y)$
+\end{enumerate}
+
+A floating-point operation causes a small rounding error:
+\begin{equation}
+    \left\Vert \frac{(x \oplus y) - (x + y)}{x+y} \right\Vert < \varepsilon_{\text{mach}}
+\end{equation}
+%
+Although, some operations may be subject to the \textbf{cancellation} problem which causes information loss.
+\begin{example}
+    Given $x = 1$ and $y = 1 \cdot 10^{-16}$, we want to compute $x + y$ in $\mathcal{F}(10, 16, U, L)$.\\
+    \begin{equation*}
+        \begin{split}
+            z & = \texttt{fl}(x) + \texttt{fl}(y) \\
+              & = 0.1 \cdot 10^1 + 0.1 \cdot 10^{-15} \\
+              & = (0.1 + 0.\overbrace{0\dots0}^{\mathclap{16\text{ zeros}}}1) \cdot 10^1 \\
+              & = 0.1\overbrace{0\dots0}^{\mathclap{15\text{ zeros}}}1 \cdot 10^1
+        \end{split}
+    \end{equation*}
+    Then, we have that $\texttt{fl}(z) = 0.1\overbrace{0\dots0}^{\mathclap{15\text{ zeros}}} \cdot 10^1 = 1 = x$.
+\end{example}
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