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+\documentclass[11pt]{beamer}
+\usetheme{Madrid}
+\usepackage[utf8]{inputenc}
+\usepackage{amsmath}
+
+\usepackage{svg}
+\usepackage{color}
+\usepackage{listings}
+\usepackage{mathtools}
+\usepackage{tikz-cd}
+\usepackage{adjustbox}
+
+\definecolor{myblue}{rgb}{0,0,0.5}
+\lstset{
+        language=Python,
+        tabsize=4,
+        basicstyle=\footnotesize,
+        keywordstyle=\bf\color{myblue},
+        commentstyle=\it\color{gray},
+        numbers=left,
+        numbersep=3pt,
+        numberstyle=\tiny\color{gray},
+}
+
+\author[\texttt{sebastiano.tronto@uni.lu}]{Sebastiano Tronto}
+\title[Students requests]%
+{Students requests}
+\logo{\includegraphics[scale=0.1]{img/unilu.jpg}} 
+%\institute{University of Luxembourg} 
+
+\date{2021-05-21} 
+
+\begin{document}
+
+\begin{frame}
+  \titlepage
+\end{frame}
+
+\begin{frame}[plain]
+        \begin{center} {\Huge More cryptography} \end{center}
+\end{frame}
+
+\begin{frame}{Cryptography}
+        What we have seen:
+
+        \vspace{0.3cm}
+        \begin{itemize}
+                \item \textbf{RSA:}
+                        sending messages using a private key / public key pair
+                \item \textbf{Flip-a-coin:}
+                        cryptographic ``proof'' that the opponent is not cheating
+        \end{itemize}
+\end{frame}
+
+\begin{frame}{Cryptography}
+        \begin{itemize}
+                \item Rely on integer factorization being hard
+
+                        \vspace{0.3cm}
+                        \textbf{Example:} the best-known factorization algorithm
+                        (\href{https://en.wikipedia.org/wiki/General\_number\_field\_sieve}%
+                        {\emph{General number field sieve}}) has complexity
+                        \begin{align*}
+                                \sim O\left(
+                                        e^{\sqrt[3]{\frac{64}{9}\log_2n\cdot(\log_2\log_2n)^2}}
+                                \right)
+                        \end{align*}
+
+                        Factoring a number with $300$ digits:
+                        \begin{itemize}
+                                \item Your laptop: $10^{13}$ billion years
+                                \item Best supercomputer: $13$ billion years
+                                        (age of the universe)
+                        \end{itemize}
+        \end{itemize}
+\end{frame}
+
+\begin{frame}{Symmetric and asymmetric cryptography}
+        \begin{itemize}
+                \item Our examples are \emph{asymmetric}: different public/private keys
+                \item Safe against eavesdroppers
+                \item Symmetric protocols can be faster and simpler, but you need
+                        a secure way to exchange a key
+        \end{itemize}
+\end{frame}
+
+\begin{frame}{Diffie-Hellman key exchange}
+        \begin{itemize}
+                \item Generate a ``password'' without communicating it directly
+                \item It can then be used for symmetric cryptography
+                \item Based on a different hard problem:
+                        \href{https://en.wikipedia.org/wiki/Discrete\_logarithm}%
+                        {\emph{discrete logarithm}}
+        \end{itemize}
+\end{frame}
+
+\begin{frame}{Diffie-Hellman key exchange}
+        \begin{itemize}
+                \item Alice and Bob agree on a prime number $p$ and an integer $g$
+                \item Alice picks an integer $a$ and sends $(g^a\bmod p)$ to Bob
+                \item Bob picks an integer $b$ and sends $(g^b\bmod p)$ to Alice
+                \item Alice can compute $(g^b)^a\bmod p$ and Bob can compute
+                        $(g^a)^b\bmod p$. This is their shared secret (key).
+        \end{itemize}
+\end{frame}
+
+\begin{frame}{Diffie-Hellman with colors (from Wikipedia)}
+        \begin{center}\includesvg[scale=0.45]{img/DH}\end{center}
+\end{frame}
+
+\begin{frame}{Diffie-Hellman key exchange}
+        \begin{itemize}
+                \item Knowing $h$ and $a$, it is hard to find $g$ such that
+                        $g^a \bmod p =h$ (discrete logarithm problem)
+                \item Very simple, many variants
+                \item Any group can be used, e.g. Elliptic Curves (see
+\href{https://en.wikipedia.org/wiki/Elliptic-curve_Diffie\%E2\%80\%93Hellman}%
+                        {Wikipedia: elliptic-curve Diffie-Hellman})
+        \end{itemize}
+\end{frame}
+
+
+\begin{frame}[plain]
+        \begin{center} {\Huge Numerical methods for PDEs} \end{center}
+\end{frame}
+
+\begin{frame}{Solving partial differential equations}
+        \begin{itemize}
+                \item Very, very hard
+                \item Very important in practical applications (physics and such)
+                \item Approximations are necessary, might as well use numerical methods
+        \end{itemize}
+\end{frame}
+
+\begin{frame}{Numerical methods for ODEs}
+        \begin{block}{Problem}
+                Given $f(x,y)$, $x_0$ and $y_0$, find an approximation
+                for $y(x)$ such that
+                \begin{align*}
+                        \begin{cases}
+                                y'(x) = f(x, y(x))\\
+                                y(x_0) =y_0
+                        \end{cases}
+                \end{align*}
+        \end{block}
+
+        \begin{block}{Approximation}
+                We can describe $y(x)$ in an interval $[x_0,x_1]$ by giving the
+                (approximate) values $y(s_0)$, \dots, $y(s_n)$ for many
+                values of $s_i\in [x_0, x_1]$.
+        \end{block}
+\end{frame}
+
+\begin{frame}{Euler's method}
+        \begin{block}{Idea}
+                For $h$ small
+                \begin{align*}
+                        y'(x)\approx\frac{y(x+h)-y(x)}{h}
+                \end{align*}
+                which implies
+                \begin{align*}
+                        y(x+h) \approx y(x) + h\cdot f(x, y(x))
+                \end{align*}
+        \end{block}
+\end{frame}
+
+\begin{frame}{Euler's method}
+        \begin{block}{Algorithm}
+                \textbf{Input:} the data $f(x,y)$, $x_0$, $y_0$ and $x_1$ describing
+                        the problem and the desired range for the solution.
+
+                \vspace{0.3cm}
+                \textbf{Output:} $x_0=s_0 < s_1 < \dots < s_n=x_1$ and
+                        $y_0, \dots, y_n$ such that $y_i\approx y(s_i)$.
+
+                \vspace{0.3cm}
+                \begin{enumerate}
+                        \item Choose a value $n$ and let
+                                $h=\frac{x_1-x_0}{n}$ and $s_i=x_0+ih$
+                        \item For $i=0,\dots, n-1$ compute
+                                $y_{i+1}=y_i+h\cdot f(s_i, y_i)$
+                        \item Return $s_0, \dots, s_n$ and $y_0, \dots, y_n$
+                \end{enumerate}
+        \end{block}
+\end{frame}
+
+\begin{frame}{Euler's method}
+        \begin{itemize}
+                \item Very simple and fast
+                \item Generalization for higher-order equations: Runge-Kutta methods
+                \item A similar idea works for some PDEs
+        \end{itemize}
+\end{frame}
+
+\begin{frame}{The heat equation (PDE)}
+        \begin{align*}
+                \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x_1^2} +
+                \frac{\partial^2 u}{\partial x_2^2} + \cdots +
+                \frac{\partial^2 u}{\partial x_n^2}
+        \end{align*}
+        
+        Where
+        \[u(x_1,x_2,\dots,x_n,t): \mathbb R^n\times \mathbb R_+\to \mathbb R\]
+        describes the quantity of heat at the point $(x_1,\dots x_n)$ at time $t$.
+
+        \vspace{0.3cm} It appears also outside thermodynamics: mathematical finance
+        (\href{https://en.wikipedia.org/wiki/Black\%E2\%80\%93Scholes\_equation}%
+        {Black-Scholes equation}), quantum mechanics
+        (\href{https://en.wikipedia.org/wiki/Schr\%C3\%B6dinger\_equation}%
+        {Schrödinger equation}), image analysis\dots
+\end{frame}
+
+\begin{frame}{A simple case ($n=1$, in $[0,1]^2$)}
+        \begin{block}{Problem}
+                Given $u_0(t)$, $u_1(t)$ and $u^0(x)$, find an approximation
+                for $u(x,t)$ such that
+                \begin{align*}
+                        \begin{cases}
+                                \frac{\partial u}{\partial t} =
+                                        \frac{\partial^2 u}{\partial x^2} \\
+                                u(0,t) = u_{(0)}(t) \quad \text{(boundary condition)}\\
+                                u(1,t) = u_{(1)}(t) \quad \text{(boundary condition)}\\
+                                u(x,0) = u^0(x) \quad \text{(initial condition)}
+                        \end{cases}
+                \end{align*}
+        \end{block}
+
+        \begin{block}{Approximation}
+                Values $u_i^j\approx u(s_i, r^j)$ for
+                $(s_i,r^j)\in [0,1]\times [0,1]$
+        \end{block}
+\end{frame}
+
+\begin{frame}{Idea}
+        For $k$ small:
+        \begin{align*}
+                \frac{\partial u(x,t)}{\partial t} \approx \frac{u(x,t+k)-u(x,t)}{k}\\
+        \end{align*}
+        For $h$ small (left limit + right limit):
+        \begin{align*}
+                \frac{\partial^2 u(x,t)}{\partial x^2} &\approx
+                        \frac{\partial}{\partial x}\left(
+                                \frac{u(x,t) - u(x-h,t)}{h}
+                        \right)\\
+                &\approx \frac1h\left(
+                        \frac{\partial u(x,t)}{\partial x} -
+                        \frac{\partial u(x-h,t)}{\partial x}
+                \right)\\
+                &\approx \frac1h\left(
+                        \frac{u(x+h,t) - u(x,t)}{h} - \frac{u(x,t)-u(x-h,t)}{h}
+                \right)\\
+                &\approx \frac{u(x+h,t)-2u(x,t)+u(x-h,t)}{h^2}
+        \end{align*}
+\end{frame}
+
+\begin{frame}{Idea}
+        From the equation
+        \begin{align*}
+                \frac{u_i^{j+1}-u_i^j}{k}= \frac{u_{i+1}^j-2u_{i}^j+u_{i-1}^j}{h^2}
+        \end{align*}
+        we find the formula
+        \begin{align*}
+                u_i^{j+1} = \frac{k}{h^2}\left(u_{i+1}^j - 2u_i^j + u_{i-1}^j\right)
+                         + u_i^j
+        \end{align*}
+\end{frame}
+
+\begin{frame}{Finite difference method for the heat equation}
+        \begin{block}{Algorithm}
+                \textbf{Input:} $u_{(0)}^j$, $u_{(1)}^j$ (boundary)
+                        and $u_i^0$ (initial).
+
+                \vspace{0.3cm}
+                \textbf{Output:} values $u_i^j$ approximating a solution.
+
+                \vspace{0.3cm}
+                \begin{enumerate}
+                        \item Let  $m=\operatorname{len}(u_0)-1$,
+                                $n=\operatorname{len}(u^0)-1$ and $k=1/m$, $h=1/n$
+                                %\begin{align*}
+                                %       \begin{array}{cccc}
+                                %               k=\frac{t_1-t_0}{m}, & h=\frac{x_1-x_0}{n}, &
+                                %               r^j = t_0 +jk, & s_i = x_0+ih
+                                %       \end{array}
+                                %\end{align*}
+                        \item For $j=0,\dots, m-1$ do the following:
+                                \begin{itemize}
+                                        \item For $i=1,\dots, n-1$ compute
+                                                \begin{align*}
+                                                        u_i^{j+1} = \frac{k}{h^2}\left(u_{i+1}^j - 
+                                                                     2u_i^j + u_{i-1}^j\right) + u_i^j
+                                                \end{align*}
+                                \end{itemize}
+                        \item Return the $u_i^j$
+                \end{enumerate}
+        \end{block}
+\end{frame}
+
+\begin{frame}{Other PDEs}
+        \begin{itemize}
+                \item In general, there is no generic method
+                \item You might need to write specific code for your equation
+                \item Some packages exists
+                        (e.g. \href{https://wiki.octave.org/Fem-fenics}{fem-fenics} for
+                        \href{https://www.gnu.org/software/octave/index}{Gnu Octave})
+        \end{itemize}
+\end{frame}
+
+\end{document}