From 08aef45f6350deec2aa7e93c12a8647bf2a64d52 Mon Sep 17 00:00:00 2001
From: Daniel Bosk <dbosk@kth.se>
Date: Tue, 5 Feb 2019 12:54:17 +0100
Subject: [PATCH 1/6] basics: breaks out contents and preamble

---
 basics/Makefile     |   3 +-
 basics/basics.tex   | 879 +-------------------------------------------
 basics/contents.tex | 855 ++++++++++++++++++++++++++++++++++++++++++
 basics/preamble.tex |  23 ++
 4 files changed, 882 insertions(+), 878 deletions(-)
 create mode 100644 basics/contents.tex
 create mode 100644 basics/preamble.tex

diff --git a/basics/Makefile b/basics/Makefile
index 9b50168..8a9f84b 100644
--- a/basics/Makefile
+++ b/basics/Makefile
@@ -4,7 +4,8 @@ all: basics.pdf
 LATEXFLAGS+= 	-shell-escape
 TEX_OUTDIR= 	.
 
-SRC+=		basics.tex abstract.tex basics.bib
+SRC+=		basics.tex preamble.tex basics.bib
+SRC+= 		abstract.tex contents.tex
 SRC+=		hhcount.sty
 FIGURES+= 	collusion.png
 FIGURES+= 	password_strength.png password_reuse.png encryptic.png
diff --git a/basics/basics.tex b/basics/basics.tex
index c49c5fb..85a1f86 100644
--- a/basics/basics.tex
+++ b/basics/basics.tex
@@ -1,27 +1,6 @@
 %\documentclass[handout]{beamer}
 \documentclass{beamer}
-\usepackage[utf8]{inputenc}
-\usepackage[T1]{fontenc}
-\usepackage[swedish,british]{babel}
-\usepackage{url}
-\usepackage{graphicx}
-\usepackage{color}
-\usepackage{subfig}
-\usepackage{multicol}
-\usepackage{amssymb,amsmath,amsthm}
-\usepackage{booktabs}
-%\usepackage[squaren,binary]{SIunits}
-\usepackage[binary-units]{siunitx}
-\usepackage[strict]{csquotes}
-\usepackage{cleveref}
-\usepackage{hhcount}
-\usepackage{pgfplots}
-
-\usepackage{mathtools}
-
-\setbeamertemplate{bibliography item}[text]
-\usepackage[natbib,style=alphabetic,maxbibnames=99]{biblatex}
-\addbibresource{basics.bib}
+\input{preamble.tex}
 
 \usepackage{xparse}
 \ProvideDocumentEnvironment{exercise}{o}{%
@@ -96,860 +75,6 @@
   \tableofcontents
 \end{frame}
 
-
-% Since this a solution template for a generic talk, very little can
-% be said about how it should be structured. However, the talk length
-% of between 15min and 45min and the theme suggest that you stick to
-% the following rules:  
-
-% - Exactly two or three sections (other than the summary).
-% - At *most* three subsections per section.
-% - Talk about 30s to 2min per frame. So there should be between about
-%   15 and 30 frames, all told.
-
-
-\section{Introduction}
-
-\subsection{History}
-
-\begin{frame}
-  \begin{itemize}
-    \item Created 1948 by Shannon's paper 
-      \citetitle{Shannon1948amt}~\cite{Shannon1948amt}.
-
-      \pause{}
-
-    \item He starts using the term \enquote{entropy} as a measure for 
-      information.
-      \begin{itemize}
-        \item In physics entropy measures the disorder of molecules.
-        \item Shannon's entropy measures disorder of information.
-      \end{itemize}
-
-      \pause{}
-
-    \item He used this theory to analyse communication.
-      \begin{itemize}
-        \item What are the theoretical limits for different channels?
-        \item How much redundancy is needed for certain noise?
-      \end{itemize}
-
-  \end{itemize}
-\end{frame}
-
-\begin{frame}
-  \begin{itemize}
-    \item This theory is interesting on the physical layer of networking.
-
-      \pause{}
-
-    \item It's also interesting for security.
-      \begin{itemize}
-        \item Field of Information Theoretic Security
-        \item \enquote{Efficiency} of passwords
-        \item Measure identifiability
-        \item \dots
-      \end{itemize}
-  \end{itemize}
-\end{frame}
-
-
-\section{Shannon entropy}
-
-\subsection{Definition of Shannon Entropy}
-
-\begin{frame}
-  \begin{definition}[Shannon entropy]
-    \begin{itemize}
-      \item Stochastic variable \(\stoch X\) assumes values from \(X\).
-      \item Shannon entropy \(H(\stoch X)\) defined as
-        \begin{align*}
-          H(\stoch X) = -K \sum_{x\in X} \Pr(\stoch X = x)\log \Pr(\stoch X = x),
-        \end{align*}
-      \item Usually \(K = \frac{1}{\log 2}\) to give entropy in unit bits  
-        (\si{\bit}).
-    \end{itemize}
-  \end{definition}
-\end{frame}
-
-\begin{frame}
-  \begin{block}{Shannon entropy can be seen as \dots}
-    \begin{itemize}
-      \item \dots how much choice in each event.
-
-      \item \dots the uncertainty of each event.
-
-      \item \dots how many bits to store each event.
-
-      \item \dots how much information it produces.
-
-    \end{itemize}
-  \end{block}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Toss a coin]
-    \begin{itemize}
-      \item Stochastic variable \(\stoch{S}\) takes values from \(S = \{h, 
-          t\}\).
-      \item We have \(\Pr(\stoch S = h) = \Pr(\stoch S = t) = \frac{1}{2}.\)
-      \item This gives \(H(\stoch S)\) as follows:
-        \begin{align*}
-          H(\stoch S) &= -\left( \Pr(\stoch S = h)\log \Pr(\stoch S = h) 
-            + \Pr(\stoch S = t) \log \Pr(\stoch S = t) \right) \\
-          &= -2\times \frac{1}{2}\log \frac{1}{2} = \log 2 = 1.
-        \end{align*}
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Roll a die]
-    \begin{itemize}
-      \item Stochastic variable \(\stoch D\) takes values from \(D 
-          = \{\fcdice{1}, \fcdice{2}, \fcdice{3}, \fcdice{4}, \fcdice{5}, 
-          \fcdice{6}\}\).
-      \item We have \(\Pr(\stoch D = d) = \frac{1}{6}\) for all \(d\in D\).
-      \item The entropy \(H(\stoch D)\) is as follows:
-        \begin{align*}
-          H(\stoch D) &= -\sum_{d\in D} \Pr(\stoch D = d)\log\Pr(\stoch D = d) \\
-          &= -6\times \frac{1}{6}\log\frac{1}{6} = \log 6 \approx 2.585.
-        \end{align*}
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item If we didn't know already, we now know that a roll of a die \dots
-        \begin{itemize}
-          \item contains more \enquote{choice} than a coin toss.
-          \item is more uncertain to predict than a coin toss.
-          \item requires more bits to store than a coin toss.
-          \item produces more information than a coin toss.
-        \end{itemize}
-
-      \item What if we modify the die a bit?
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Roll of a modified die]
-    \begin{itemize}
-      \item Stochastic variable \(D'\) taking values from \(D\).
-      \item We now have \(\Pr(\stoch D' = \fcdice{6}) = \frac{9}{10}\) and 
-        \(\Pr(\stoch D' = d) = \frac{1}{10}\times\frac{1}{5}\) for \(d\neq 
-          \fcdice{6}\).
-      \item This yields
-        \begin{align*}
-          H(\stoch D') &= -\left( \frac{9}{10}\log\frac{9}{10} + \sum_{d\neq 6} 
-            \frac{1}{50}\log\frac{1}{50} \right) \\
-          &= -\frac{9}{10}\log\frac{9}{10} -5\times\frac{1}{50}\log\frac{1}{50} 
-          \\
-          &= -\frac{9}{10}\log\frac{9}{10} -\frac{1}{10}\log\frac{1}{50} 
-          \approx 0.701.
-        \end{align*}
-      \item Note that the \(\log\) function is the logarithm in base 2 (i.e.\ 
-        \(\log_2\)).
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item This die is much easier to predict.
-      \item It produces much less information --- less than a coin toss!
-      \item Requires less data for storage etc.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\subsection{Properties for Shannon entropy}
-
-\begin{frame}
-  \begin{definition}
-    \begin{itemize}
-      \item Function \(f\colon \R\to \R\) such that
-        \begin{align*}
-          tf(x) + (1-t)f(y) \leq f(tx + (1-t)y),
-        \end{align*}
-
-      \item Then \(f\) is \emph{concave}.
-      \item With strict inequality for \(x\neq y\) we say that \(f\) is 
-        \emph{strictly concave}.
-    \end{itemize}
-  \end{definition}
-
-  \begin{example}
-    \(\log\colon \R\to \R\) is strictly concave.
-  \end{example}
-\end{frame}
-
-\begin{frame}[fragile]
-  \begin{tikzpicture}
-    \begin{axis}[xlabel=$x$,xmin=0.5,ylabel=$\log x$]
-      \addplot gnuplot[id=log]{log(x)};
-    \end{axis}
-  \end{tikzpicture}
-\end{frame}
-
-%\begin{frame}
-%  \begin{lemma}
-%    Låt \(f\) vara en strikt konkav funktion.
-%    Då har vi att
-%    \begin{align*}
-%      tf(x) + (1-y)f(y) = f( tx + (1-t)y )
-%    \end{align*}
-%    om och endast om \(x = y\).
-%  \end{lemma}
-%\end{frame}
-%
-%\begin{frame}
-%  \begin{proof}
-%    Antag \(x = y\).
-%    Då har vi
-%    \begin{align*}
-%      tf(x) + (1-t)f(x) = f(x)(t+1-t) = f(x).
-%    \end{align*}
-%    Men
-%    \begin{align*}
-%      f(tx + (1-t)x) = f((t+1-t)x) = f(x)
-%    \end{align*}
-%    och alltså har vi likhet.
-%
-%    Antag \(tf(x) + (1-t)f(y) = f(tx + (1-t)y)\).
-%    Då har vi
-%    \begin{align*}
-%      t( f(x) - f(y) ) + f(y) = f( t( x - y ) + y ).
-%    \end{align*}
-%    Högerledet implicerar \(f(x) = f(y)\), men då beror vänterledet enbart på 
-%    \(f(y)\) och således måste även \(x = y\).
-%  \end{proof}
-%\end{frame}
-
-\begin{frame}
-  \begin{theorem}[Jensen's inequality]
-    \begin{itemize}
-      \item Strictly concave function \(f\colon \R\to \R\).
-      \item Real numbers \(a_1, a_2,\ldots, a_n > 0\) such that \(\sum_{i=1}^n 
-          a_i = 1\).
-      \item Then we have
-        \begin{align*}
-          \sum_{i=1}^n a_i f(x_i) \leq f\left( \sum_{i=1}^n a_i x_i\right).
-        \end{align*}
-      \item We have equality iff \(x_1 = x_2 = \cdots = x_n\).
-    \end{itemize}
-  \end{theorem}
-\end{frame}
-
-%\begin{frame}
-%  \begin{block}{Bevis för Jensens olikhet.}
-%    Bevis genom induktion.
-%    Antag \(n=2\).
-%    Då har vi att \(a_1 + a_2 = 1\) och alltså \(a_1 = 1 - a_2\).
-%    Eftersom \(f\) är konkav har vi att
-%    \begin{align*}
-%      a_1f(x_1) + a_2f(x_2) \leq f(a_1x_1 + a_2x_2).
-%    \end{align*}
-%
-%    Antag sant för \(n=k\).
-%    \begin{align}
-%      \label{eq:JensenIndhyp}
-%      \sum_{i=1}^k a_i = 1 \land
-%        \sum_{i=1}^k a_if(x_i) \leq f\left( \sum_{i=1}^k a_ix_i\right).
-%    \end{align}
-%  \end{block}
-%\end{frame}
-%
-%\begin{frame}
-%  \begin{proof}[Forts. bevis för Jensens olikhet]
-%    Lås oss visa att detta även gäller för \(n=k+1\).
-%    Eftersom att \(f\) är konkav gäller att
-%    \begin{align}
-%      \label{eq:JensenInductive1}
-%      \sum_{i=1}^{k+1} a_if(x_i) &= a_1f(x_1) + (1-a_1)\sum_{i=2}^{k+1} 
-%      \frac{a_i}{1-a_1}f(x_i) \\
-%      \label{eq:JensenInductive2}
-%        &\leq f\left(a_1x_1 + (1-a_i)\sum_{i=2}^{k+1} \frac{a_i}{1-a_1} 
-%        x_i\right).
-%    \end{align}
-%    Då \(\sum_{i=2}^{k+1}\frac{a_i}{1-a_i} = 1\) kan vi tillämpa 
-%    induktionshypotesen \eqref{eq:JensenIndhyp} i \eqref{eq:JensenInductive1} 
-%    och \eqref{eq:JensenInductive2}, alltså är det sant för alla \(n\in \N\).
-%
-%    Vidare följer likhet från lemmat ovan.
-%  \end{proof}
-%\end{frame}
-
-\begin{frame}
-  \begin{theorem}
-    \begin{itemize}
-      \item Stochastic variable \(\stoch X\) with probability distribution 
-        \begin{equation*}
-          p_1, p_2,\ldots, p_n, \text{ where } p_i > 0 \text{ for } 1\leq i\leq 
-          n.
-        \end{equation*}
-      \item Then \(H(\stoch X)\leq \log n\).
-      \item Equality iff \(p_1 = p_2 = \cdots = p_n = 1/n\).
-    \end{itemize}
-  \end{theorem}
-\end{frame}
-
-\begin{frame}
-  \begin{proof}
-    The theorem follows directly from Jensen's inequality:
-    \begin{align*}
-      H(\stoch X) &= -\sum_{i=1}^n p_i\log p_i = \sum_{i=1}^n 
-      p_i\log\frac{1}{p_i} \\
-      &\leq \log\sum_{i=1}^n p_i\frac{1}{p_i} = \log n.
-    \end{align*}
-    With equality iff \(p_1 = p_2 = \cdots = p_n\).
-  \end{proof}
-\end{frame}
-
-\begin{frame}
-  \begin{corollary}
-    \(H(\stoch X) = 0\) iff \(\Pr(\stoch X = x) = 1\) for some \(x\in X\) and 
-    \(\Pr(\stoch X = x^\prime) = 0\) for all \(x\neq x^\prime \in X\).
-  \end{corollary}
-
-  \begin{proof}
-    \begin{itemize}
-      \item If \(\Pr(\stoch X = x) = 1\), then \(n = 1\) and thus \(H(\stoch X) 
-          = \log n = 0\).
-
-      \item If \(H(\stoch X) = 0\), then \(H(\stoch X) \leq \log n = 0\).
-        Thus \(n = 1\).
-    \end{itemize}
-  \end{proof}
-\end{frame}
-
-%\begin{frame}
-%  \begin{theorem}
-%    Följande egenskaper gäller:
-%    \begin{enumerate}
-%      \item\label{prop:cont} \(H\) är kontinuerlig.
-%      \item\label{prop:mono} Om \(\Pr(\stoch X = x) = 1/|X|\) för alla \(x\in 
-%        X\) då är \(H\) en monotont stigande funktion med avseende på \(|X|\).
-%    \end{enumerate}
-%  \end{theorem}
-%
-%  \begin{proof}
-%    \Cref{prop:cont} följer direkt av att logaritmen är kontinuerlig och 
-%    funktionssammansättningar av kontinuerliga funktioner är kontinuerliga.
-%
-%    \Cref{prop:mono} följer av föregående sats.
-%  \end{proof}
-%\end{frame}
-
-\begin{frame}
-  \begin{lemma}
-    \begin{itemize}
-      \item Stochastic variables \(\stoch X\) and \(\stoch Y\).
-      \item Then we have
-        \begin{align*}
-          H(\stoch X, \stoch Y)\leq H(\stoch X) + H(\stoch Y).
-        \end{align*}
-      \item Equality iff \(\stoch X\) and \(\stoch Y\) are independent.
-    \end{itemize}
-  \end{lemma}
-\end{frame}
-
-%\begin{frame}
-%  \begin{proof}
-%  \end{proof}
-%\end{frame}
-%
-%\begin{frame}
-%  \begin{theorem}
-%    En utjämning av sannolikheterna ökar \(H(\stoch X)\).
-%  \end{theorem}
-%\end{frame}
-%
-%\begin{frame}
-%  \begin{proof}
-%  \end{proof}
-%\end{frame}
-
-\subsection{Conditional entropy}
-
-\begin{frame}
-  \begin{definition}[Conditional entropy]
-    \begin{itemize}
-      \item Define \emph{conditional entropy} \(H(\stoch Y\mid \stoch X)\) 
-        as
-        \begin{align*}
-          H(\stoch Y\mid \stoch X) = %-\sum_{i,j} \Pr(i,j)\log \Pr(j\mid i)
-          -\sum_y\sum_x \Pr(\stoch Y = y)\Pr(\stoch X = x\mid y)\log \Pr(\stoch 
-          X = x\mid y).
-        \end{align*}
-    \end{itemize}
-  \end{definition}
-
-  \pause{}
-
-  \begin{remark}
-    This is the uncertainty in \(\stoch Y\) which is not revealed by \(\stoch 
-      X\).
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{theorem}
-    \(H(\stoch X, \stoch Y) = {\color{red}H(\stoch X)} + {\color{green}H(\stoch 
-        Y\mid \stoch X)}\)
-  \end{theorem}
-
-  \begin{center}
-  \begin{tikzpicture}
-    \def\HX{(0,0) circle (2)}
-    \def\HY{(2,0) circle (2)}
-    \def\HYX{(1,0)}
-    \begin{scope}%[fill opacity=0.75]
-      \fill[green] \HY;
-      \fill[red] \HX;
-      \draw \HX node[left] {$H(X)$};
-      \draw \HY node[right] {$H(Y\mid X)$};
-    \end{scope}
-  \end{tikzpicture}
-  \end{center}
-\end{frame}
-
-%\begin{frame}
-%  \begin{proof}
-%  \end{proof}
-%\end{frame}
-
-\begin{frame}
-  \begin{corollary}
-    \(H(\stoch X\mid \stoch Y) \leq H(\stoch X)\).
-  \end{corollary}
-
-  \begin{corollary}
-    \(H(\stoch X\mid \stoch Y) = H(\stoch X)\) iff \(\stoch X\) and \(\stoch 
-      Y\) independent.
-  \end{corollary}
-\end{frame}
-%
-%\begin{frame}
-%  \begin{proof}
-%  \end{proof}
-%\end{frame}
-%
-%\begin{frame}
-%  \begin{theorem}
-%    Entropin för en Markovprocess
-%  \end{theorem}
-%\end{frame}
-
-\subsection{Information density and redundancy}
-
-\begin{frame}
-  \begin{definition}
-    \begin{itemize}
-      \item Natural language \(L\).
-      \item Stochastic variable \(\stoch P^n_L\) of strings of length \(n\).
-      \item (Alphabet \(P_L\).)
-      \item Entropy of \(L\) defined as
-        \begin{align*}
-          H_L = \lim_{n\to \infty}\frac{H(\stoch P^n_L)}{n}.
-        \end{align*}
-      \item Redundancy in \(L\) is
-        \begin{align*}
-          R_L = 1 - \frac{H_L}{\log |P_L|}.
-        \end{align*}
-    \end{itemize}
-  \end{definition}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    Meaning we have \(H_L\) bits per character in \(L\).
-  \end{remark}
-
-  \begin{example}[\cite{Shannon1948amt}]
-    \begin{itemize}
-      \item Entropy of 1--1.5 bits per character in English.
-      \item Redundancy of approximately \(1 - \frac{1.25}{\log 26} \approx 
-          0.73\).
-    \end{itemize}
-  \end{example}
-
-\end{frame}
-
-\begin{frame}
-  \begin{example}[\cite{Shannon1948amt}]
-    Two-dimensional cross-word puzzles requires redundancy of approximately 
-    \(0.5\).
-  \end{example}
-
-  \begin{example}
-    \begin{itemize}
-      \item Redundancy of \enquote{SMS languages} is lower than for 
-        \enquote{non-SMS languages}.
-
-      \item Compare \enquote{också} and \enquote{oxå}.
-
-    \end{itemize}
-  \end{example}
-
-  \begin{remark}
-    \begin{itemize}
-      \item Lower redundancy is more space-efficient.
-      \item Incurs more errors.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-%\begin{frame}
-%  \begin{itemize}
-%    \item Detta säger också att vi kan uppskatta entropin för en given 
-%      Markovprocess.
-%
-%    \item Shannon modellerade språket som en Markovprocess i sin artikel 
-%      \cite{Shannon1948amt}.
-%
-%    \item Vi kan även beräkna entropin för ett givet tillstånd i en 
-%      Markovprocess genom betingad entropi.
-%
-%  \end{itemize}
-%\end{frame}
-
-\subsection{Information gain}
-
-\begin{frame}
-  \begin{definition}
-    \begin{itemize}
-      \item Set \(U\) of possible outcomes.
-      \item Probability of outcome \(u\in U\) denoted \(p_u\).
-      \item We learn that some \emph{unknown} outcome is in \(A\subset U\).
-      \item Then the \emph{information gain} \(G(A\mid U)\) is defined as
-        \begin{align*}
-          G(A\mid U) = \log\frac{1}{\Pr(A)} = -\log\Pr(A),
-        \end{align*}
-        where \(\Pr(A) = \sum_{i\in A} p_i\).
-    \end{itemize}
-  \end{definition}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Roll of dice again]
-    \begin{itemize}
-      \item Someone rolls and we should guess the result, \(\frac{1}{6}\) 
-        chance.
-      \item We learn that it was an even number, we gain
-        \begin{align*}
-          -\log\left( \frac{1}{6} + \frac{1}{6} + \frac{1}{6}\right) =
-          -\log\frac{3}{6} = \log\frac{6}{3} = \log 2 = 1.
-        \end{align*}
-      \item The remaining uncertainty is \SI{1.58}{\bit}.
-    \end{itemize}
-  \end{example}
-
-  \pause{}
-
-  \begin{remark}
-    \begin{itemize}
-      \item \(X' = \{\fcdice{2}, \fcdice{4}, \fcdice{6}\}\)
-      \item \(H(\stoch X') = - \sum_{x\in X'} \Pr(\stoch X' = x)\log \Pr(\stoch 
-          X' = x)\)
-      \item I.e.\ \(- 3 \times \frac{1}{3}\log\frac{1}{3} \approx 1.58\).
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Dice yet again]
-    \begin{itemize}
-      \item We learn the die show less than five, i.e.\ not \fcdice{5} nor 
-        \fcdice{6}.
-      \item This yields
-        \begin{align*}
-          -\log\left( 4\times\frac{1}{6}\right) = \log\frac{6}{4}\approx 0.58
-        \end{align*}
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-
-\section[Applications]{Application in security}
-
-\subsection{Passwords}
-
-\begin{frame}
-  \begin{block}{Idea~\cite{Komanduri2011opa}}
-    \begin{itemize}
-      \item Look at different aspects of passwords individually, then 
-        summarize.
-      \item Can use \(H(x_1, x_2, \ldots, x_n) \leq H(x_1) + H(x_2) + \cdots 
-          + H(x_n)\).
-      \item This allows us to reason about bounds.
-    \end{itemize}
-  \end{block}
-\end{frame}
-
-\begin{frame}
-  \begin{example}
-    \begin{itemize}
-      \item We can look at properties such as:
-        \begin{itemize}
-          \item length,
-          \item number of and placement of character classes,
-          \item the actual characters,
-          \item \dots
-        \end{itemize}
-    \end{itemize}
-  \end{example}
-
-  \pause{}
-
-  \begin{remark}
-    \begin{itemize}
-      \item These are \emph{not independent}.
-      \item The sum will be an \emph{upper bound}.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item With an upper bound we know it's not possible to do better.
-      \item With an average we know how well most users will do.
-      \item With a lower bound we have a guarantee --- not possible!
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item If a password policy yields low entropy, it implies it's bad.
-      \item If a password policy yields high entropy, it \emph{doesn't} imply 
-        that it's good.
-    \end{itemize}
-  \end{remark}
-
-  \pause
-
-  \begin{exercise}
-    Why?
-  \end{exercise}
-\end{frame}
-
-\begin{frame}
-  \begin{figure}
-    \includegraphics[height=0.7\textheight]{password_strength.png}
-    \caption{xkcd's strip on password strength.
-    Picture: xkcd~\cite{xkcd936}.}
-  \end{figure}
-\end{frame}
-
-%\begin{frame}{En förklaring av xkcd}
-%  \begin{itemize}
-%    \item Vi har 1 miljon engelska ord: ger \(\log 10^6 \approx 20\) bitar 
-%      entropi.
-%      (xkcd använder 16 bitar, vilket ger ca 70\,000 ord, alla ord i engelskan 
-%      är inte vanliga.)
-%
-%    \item Vi kan ha inledande versal: ger 1 bit entropi.
-%
-%    \item Vi har några vanliga substitutioner: uppskattningsvis 10 stycken, 
-%      d.v.s.~3 bitar entropi.
-%
-%    \item Vi har specialtecken (ej substitution): uppskattningsvis 4 bitar 
-%      entropi.
-%
-%    \item Vi har siffror: \(\log 10\approx 3\).
-%
-%    \item Ordningen på specialtecknet och siffran: ger 1 bit entropi.
-%
-%    \item Totalt 32 bitar entropi:
-%      \begin{itemize}
-%        \item Tar minst 50 dagar med 1\,000 gissningar per sekund.
-%        \item Tar strax över en timme med 1\,000\,000 gissningar per sekund.
-%      \end{itemize}
-%
-%  \end{itemize}
-%\end{frame}
-
-\begin{frame}
-  \begin{example}[Standard password]
-    \begin{itemize}
-      \item We have
-        \begin{itemize}
-          \item 26 alphabetic characters,
-          \item 10 numbers,
-          \item 10 special characters (approximately).
-        \end{itemize}
-
-      \item This yields \(\log( 2\times 26 + 10 + 10 ) = \log 72 \approx 
-          \SI{6}{\bit}\) per password character.
-
-      \item A 10-character \emph{uniformly randomly} generated password 
-        contains \SI{60}{\bit}.
-    \end{itemize}
-  \end{example}
-
-  \pause{}
-
-  \begin{remark}
-    What happens when we require two upper and two lower-case characters, two 
-    numbers must be included?
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Four-word passphrase]
-    \begin{itemize}
-      \item We have 125\,000 words in the standard Swedish dictionary.
-      \item This yields \(\log 125\,000\approx \SI{17}{\bit}\) per word.
-      \item A four-word \emph{uniformly randomly} generated passphrase contains
-        \SI{68}{\bit}.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Random sentence]
-    \begin{itemize}
-      \item We estimated the entropy per character in a language.
-      \item It was approximately \(\SI{1.25}{\bit}\) for English.
-      \item A 20-character \emph{uniformly randomly} generated sentence would 
-        yield \SI{25}{\bit}.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item All these require uniform randomness.
-      \item Humans are bad at remembering random things.
-      \item Thus they will choose non-randomly.
-      \item The entropy will thus be (possibly much) lower.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\subsection{Research about human chosen passwords}
-
-\begin{frame}
-  \begin{example}[\citetitle{Bonneau2012lpo}~\cite{Bonneau2012lpo}]
-    \begin{itemize}
-      \item Investigates how linguistics affect the choice of multi-word 
-        passphrases.
-
-      \item Users don't choose them randomly, prefer adapted to natural 
-        language.
-
-      \item \enquote{correct horse battery staple} is preferred to 
-        \enquote{horse correct battery staple} since the first is more 
-        grammatically correct.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[\citetitle{Kuo2006hso}~\cite{Kuo2006hso}]
-    \begin{itemize}
-      \item Studied how users creates easy-to-remember passwords.
-
-      \item Also investigated the strength of phrase-based passwords.
-
-      \item E.g.\ Google's example \enquote{To be or not to be, that is the 
-          question}\footnote{%
-          URL\@: 
-          \protect\url{http://www.lightbluetouchpaper.org/2011/11/08/want-to-create-a-really-strong-password-dont-ask-google/}.
-        } which results in \enquote{2bon2btitq}.
-
-      \item This particular password has apparently been used by many \dots
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item There is a PhD thesis on the topic of guessing passwords: 
-        \fullcite{GuessingHumanChosenSecrets}.
-      \item There is even a conference dedicated to passwords: PasswordsCon.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\subsection{Identifying information}
-
-\begin{frame}
-  \begin{example}
-    Do we get more information from zodiac signs or birthdays?
-    \begin{align*}
-      -\sum_{\mathclap{\text{zodiacs}}} \frac{1}{12} \log\frac{1}{12} &= \log 12 
-      \approx 3.58 \\
-      &< -\sum_{\mathclap{\text{days of year}}} \frac{1}{365} \log\frac{1}{365} 
-      = \log 365 \approx 8.51.
-    \end{align*}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{exercise}
-    How much information do we need to uniquely identify an individual?
-  \end{exercise}
-\end{frame}
-
-\begin{frame}
-  \begin{example}
-    \begin{itemize}
-      \item Sometime during 2011 there were \(n = 6\,973\,738\,433\)\footnote{%
-          According to the World Bank.
-        } people on earth.
-
-      \item To give everyone a unique identifier we need \(\log n\approx 
-          32.7\approx 33\) bits of information.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{block}{Identifying information in browsers}
-    \begin{itemize}
-      \item Electronic Frontier Foundation (EFF) studied~\cite{Eckersley2010hui} 
-        how much information a web-browser shares.
-
-      \item You can try your browser in \url{http://panopticlick.eff.org/}.
-    \end{itemize}
-  \end{block}
-
-  \pause{}
-
-  \begin{example}[My browser]
-    \begin{itemize}
-      \item My Firefox-browser with all addons gave 21.45 bits of entropy.
-
-      \item Then the number of tested users were 2\,860\,696.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{figure}
-    \includegraphics[height=0.7\textheight]{collusion.png}
-    \caption{Screenshot from Collusion (now Lightbeam) for Firefox.
-      Map over all pages that track me using this information.}
-  \end{figure}
-\end{frame}
-
-
-%%%%%%%%%%%%%%%%%%%%%%
-
-\subsection*{References}
-\begin{frame}[allowframebreaks]
-	\small
-  \printbibliography{}
-\end{frame}
+\mode<all>{\input{contents.tex}}
 
 \end{document}
diff --git a/basics/contents.tex b/basics/contents.tex
new file mode 100644
index 0000000..7f722e6
--- /dev/null
+++ b/basics/contents.tex
@@ -0,0 +1,855 @@
+% Since this a solution template for a generic talk, very little can
+% be said about how it should be structured. However, the talk length
+% of between 15min and 45min and the theme suggest that you stick to
+% the following rules:  
+
+% - Exactly two or three sections (other than the summary).
+% - At *most* three subsections per section.
+% - Talk about 30s to 2min per frame. So there should be between about
+%   15 and 30 frames, all told.
+
+
+\section{Introduction}
+
+\subsection{History}
+
+\begin{frame}
+  \begin{itemize}
+    \item Created 1948 by Shannon's paper 
+      \citetitle{Shannon1948amt}~\cite{Shannon1948amt}.
+
+      \pause{}
+
+    \item He starts using the term \enquote{entropy} as a measure for 
+      information.
+      \begin{itemize}
+        \item In physics entropy measures the disorder of molecules.
+        \item Shannon's entropy measures disorder of information.
+      \end{itemize}
+
+      \pause{}
+
+    \item He used this theory to analyse communication.
+      \begin{itemize}
+        \item What are the theoretical limits for different channels?
+        \item How much redundancy is needed for certain noise?
+      \end{itemize}
+
+  \end{itemize}
+\end{frame}
+
+\begin{frame}
+  \begin{itemize}
+    \item This theory is interesting on the physical layer of networking.
+
+      \pause{}
+
+    \item It's also interesting for security.
+      \begin{itemize}
+        \item Field of Information Theoretic Security
+        \item \enquote{Efficiency} of passwords
+        \item Measure identifiability
+        \item \dots
+      \end{itemize}
+  \end{itemize}
+\end{frame}
+
+
+\section{Shannon entropy}
+
+\subsection{Definition of Shannon Entropy}
+
+\begin{frame}
+  \begin{definition}[Shannon entropy]
+    \begin{itemize}
+      \item Stochastic variable \(\stoch X\) assumes values from \(X\).
+      \item Shannon entropy \(H(\stoch X)\) defined as
+        \begin{align*}
+          H(\stoch X) = -K \sum_{x\in X} \Pr(\stoch X = x)\log \Pr(\stoch X = x),
+        \end{align*}
+      \item Usually \(K = \frac{1}{\log 2}\) to give entropy in unit bits  
+        (\si{\bit}).
+    \end{itemize}
+  \end{definition}
+\end{frame}
+
+\begin{frame}
+  \begin{block}{Shannon entropy can be seen as \dots}
+    \begin{itemize}
+      \item \dots how much choice in each event.
+
+      \item \dots the uncertainty of each event.
+
+      \item \dots how many bits to store each event.
+
+      \item \dots how much information it produces.
+
+    \end{itemize}
+  \end{block}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Toss a coin]
+    \begin{itemize}
+      \item Stochastic variable \(\stoch{S}\) takes values from \(S = \{h, 
+          t\}\).
+      \item We have \(\Pr(\stoch S = h) = \Pr(\stoch S = t) = \frac{1}{2}.\)
+      \item This gives \(H(\stoch S)\) as follows:
+        \begin{align*}
+          H(\stoch S) &= -\left( \Pr(\stoch S = h)\log \Pr(\stoch S = h) 
+            + \Pr(\stoch S = t) \log \Pr(\stoch S = t) \right) \\
+          &= -2\times \frac{1}{2}\log \frac{1}{2} = \log 2 = 1.
+        \end{align*}
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Roll a die]
+    \begin{itemize}
+      \item Stochastic variable \(\stoch D\) takes values from \(D 
+          = \{\fcdice{1}, \fcdice{2}, \fcdice{3}, \fcdice{4}, \fcdice{5}, 
+          \fcdice{6}\}\).
+      \item We have \(\Pr(\stoch D = d) = \frac{1}{6}\) for all \(d\in D\).
+      \item The entropy \(H(\stoch D)\) is as follows:
+        \begin{align*}
+          H(\stoch D) &= -\sum_{d\in D} \Pr(\stoch D = d)\log\Pr(\stoch D = d) \\
+          &= -6\times \frac{1}{6}\log\frac{1}{6} = \log 6 \approx 2.585.
+        \end{align*}
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item If we didn't know already, we now know that a roll of a die \dots
+        \begin{itemize}
+          \item contains more \enquote{choice} than a coin toss.
+          \item is more uncertain to predict than a coin toss.
+          \item requires more bits to store than a coin toss.
+          \item produces more information than a coin toss.
+        \end{itemize}
+
+      \item What if we modify the die a bit?
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Roll of a modified die]
+    \begin{itemize}
+      \item Stochastic variable \(D'\) taking values from \(D\).
+      \item We now have \(\Pr(\stoch D' = \fcdice{6}) = \frac{9}{10}\) and 
+        \(\Pr(\stoch D' = d) = \frac{1}{10}\times\frac{1}{5}\) for \(d\neq 
+          \fcdice{6}\).
+      \item This yields
+        \begin{align*}
+          H(\stoch D') &= -\left( \frac{9}{10}\log\frac{9}{10} + \sum_{d\neq 6} 
+            \frac{1}{50}\log\frac{1}{50} \right) \\
+          &= -\frac{9}{10}\log\frac{9}{10} -5\times\frac{1}{50}\log\frac{1}{50} 
+          \\
+          &= -\frac{9}{10}\log\frac{9}{10} -\frac{1}{10}\log\frac{1}{50} 
+          \approx 0.701.
+        \end{align*}
+      \item Note that the \(\log\) function is the logarithm in base 2 (i.e.\ 
+        \(\log_2\)).
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item This die is much easier to predict.
+      \item It produces much less information --- less than a coin toss!
+      \item Requires less data for storage etc.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\subsection{Properties for Shannon entropy}
+
+\begin{frame}
+  \begin{definition}
+    \begin{itemize}
+      \item Function \(f\colon \R\to \R\) such that
+        \begin{align*}
+          tf(x) + (1-t)f(y) \leq f(tx + (1-t)y),
+        \end{align*}
+
+      \item Then \(f\) is \emph{concave}.
+      \item With strict inequality for \(x\neq y\) we say that \(f\) is 
+        \emph{strictly concave}.
+    \end{itemize}
+  \end{definition}
+
+  \begin{example}
+    \(\log\colon \R\to \R\) is strictly concave.
+  \end{example}
+\end{frame}
+
+\begin{frame}[fragile]
+  \begin{tikzpicture}
+    \begin{axis}[xlabel=$x$,xmin=0.5,ylabel=$\log x$]
+      \addplot gnuplot[id=log]{log(x)};
+    \end{axis}
+  \end{tikzpicture}
+\end{frame}
+
+%\begin{frame}
+%  \begin{lemma}
+%    Låt \(f\) vara en strikt konkav funktion.
+%    Då har vi att
+%    \begin{align*}
+%      tf(x) + (1-y)f(y) = f( tx + (1-t)y )
+%    \end{align*}
+%    om och endast om \(x = y\).
+%  \end{lemma}
+%\end{frame}
+%
+%\begin{frame}
+%  \begin{proof}
+%    Antag \(x = y\).
+%    Då har vi
+%    \begin{align*}
+%      tf(x) + (1-t)f(x) = f(x)(t+1-t) = f(x).
+%    \end{align*}
+%    Men
+%    \begin{align*}
+%      f(tx + (1-t)x) = f((t+1-t)x) = f(x)
+%    \end{align*}
+%    och alltså har vi likhet.
+%
+%    Antag \(tf(x) + (1-t)f(y) = f(tx + (1-t)y)\).
+%    Då har vi
+%    \begin{align*}
+%      t( f(x) - f(y) ) + f(y) = f( t( x - y ) + y ).
+%    \end{align*}
+%    Högerledet implicerar \(f(x) = f(y)\), men då beror vänterledet enbart på 
+%    \(f(y)\) och således måste även \(x = y\).
+%  \end{proof}
+%\end{frame}
+
+\begin{frame}
+  \begin{theorem}[Jensen's inequality]
+    \begin{itemize}
+      \item Strictly concave function \(f\colon \R\to \R\).
+      \item Real numbers \(a_1, a_2,\ldots, a_n > 0\) such that \(\sum_{i=1}^n 
+          a_i = 1\).
+      \item Then we have
+        \begin{align*}
+          \sum_{i=1}^n a_i f(x_i) \leq f\left( \sum_{i=1}^n a_i x_i\right).
+        \end{align*}
+      \item We have equality iff \(x_1 = x_2 = \cdots = x_n\).
+    \end{itemize}
+  \end{theorem}
+\end{frame}
+
+%\begin{frame}
+%  \begin{block}{Bevis för Jensens olikhet.}
+%    Bevis genom induktion.
+%    Antag \(n=2\).
+%    Då har vi att \(a_1 + a_2 = 1\) och alltså \(a_1 = 1 - a_2\).
+%    Eftersom \(f\) är konkav har vi att
+%    \begin{align*}
+%      a_1f(x_1) + a_2f(x_2) \leq f(a_1x_1 + a_2x_2).
+%    \end{align*}
+%
+%    Antag sant för \(n=k\).
+%    \begin{align}
+%      \label{eq:JensenIndhyp}
+%      \sum_{i=1}^k a_i = 1 \land
+%        \sum_{i=1}^k a_if(x_i) \leq f\left( \sum_{i=1}^k a_ix_i\right).
+%    \end{align}
+%  \end{block}
+%\end{frame}
+%
+%\begin{frame}
+%  \begin{proof}[Forts. bevis för Jensens olikhet]
+%    Lås oss visa att detta även gäller för \(n=k+1\).
+%    Eftersom att \(f\) är konkav gäller att
+%    \begin{align}
+%      \label{eq:JensenInductive1}
+%      \sum_{i=1}^{k+1} a_if(x_i) &= a_1f(x_1) + (1-a_1)\sum_{i=2}^{k+1} 
+%      \frac{a_i}{1-a_1}f(x_i) \\
+%      \label{eq:JensenInductive2}
+%        &\leq f\left(a_1x_1 + (1-a_i)\sum_{i=2}^{k+1} \frac{a_i}{1-a_1} 
+%        x_i\right).
+%    \end{align}
+%    Då \(\sum_{i=2}^{k+1}\frac{a_i}{1-a_i} = 1\) kan vi tillämpa 
+%    induktionshypotesen \eqref{eq:JensenIndhyp} i \eqref{eq:JensenInductive1} 
+%    och \eqref{eq:JensenInductive2}, alltså är det sant för alla \(n\in \N\).
+%
+%    Vidare följer likhet från lemmat ovan.
+%  \end{proof}
+%\end{frame}
+
+\begin{frame}
+  \begin{theorem}
+    \begin{itemize}
+      \item Stochastic variable \(\stoch X\) with probability distribution 
+        \begin{equation*}
+          p_1, p_2,\ldots, p_n, \text{ where } p_i > 0 \text{ for } 1\leq i\leq 
+          n.
+        \end{equation*}
+      \item Then \(H(\stoch X)\leq \log n\).
+      \item Equality iff \(p_1 = p_2 = \cdots = p_n = 1/n\).
+    \end{itemize}
+  \end{theorem}
+\end{frame}
+
+\begin{frame}
+  \begin{proof}
+    The theorem follows directly from Jensen's inequality:
+    \begin{align*}
+      H(\stoch X) &= -\sum_{i=1}^n p_i\log p_i = \sum_{i=1}^n 
+      p_i\log\frac{1}{p_i} \\
+      &\leq \log\sum_{i=1}^n p_i\frac{1}{p_i} = \log n.
+    \end{align*}
+    With equality iff \(p_1 = p_2 = \cdots = p_n\).
+  \end{proof}
+\end{frame}
+
+\begin{frame}
+  \begin{corollary}
+    \(H(\stoch X) = 0\) iff \(\Pr(\stoch X = x) = 1\) for some \(x\in X\) and 
+    \(\Pr(\stoch X = x^\prime) = 0\) for all \(x\neq x^\prime \in X\).
+  \end{corollary}
+
+  \begin{proof}
+    \begin{itemize}
+      \item If \(\Pr(\stoch X = x) = 1\), then \(n = 1\) and thus \(H(\stoch X) 
+          = \log n = 0\).
+
+      \item If \(H(\stoch X) = 0\), then \(H(\stoch X) \leq \log n = 0\).
+        Thus \(n = 1\).
+    \end{itemize}
+  \end{proof}
+\end{frame}
+
+%\begin{frame}
+%  \begin{theorem}
+%    Följande egenskaper gäller:
+%    \begin{enumerate}
+%      \item\label{prop:cont} \(H\) är kontinuerlig.
+%      \item\label{prop:mono} Om \(\Pr(\stoch X = x) = 1/|X|\) för alla \(x\in 
+%        X\) då är \(H\) en monotont stigande funktion med avseende på \(|X|\).
+%    \end{enumerate}
+%  \end{theorem}
+%
+%  \begin{proof}
+%    \Cref{prop:cont} följer direkt av att logaritmen är kontinuerlig och 
+%    funktionssammansättningar av kontinuerliga funktioner är kontinuerliga.
+%
+%    \Cref{prop:mono} följer av föregående sats.
+%  \end{proof}
+%\end{frame}
+
+\begin{frame}
+  \begin{lemma}
+    \begin{itemize}
+      \item Stochastic variables \(\stoch X\) and \(\stoch Y\).
+      \item Then we have
+        \begin{align*}
+          H(\stoch X, \stoch Y)\leq H(\stoch X) + H(\stoch Y).
+        \end{align*}
+      \item Equality iff \(\stoch X\) and \(\stoch Y\) are independent.
+    \end{itemize}
+  \end{lemma}
+\end{frame}
+
+%\begin{frame}
+%  \begin{proof}
+%  \end{proof}
+%\end{frame}
+%
+%\begin{frame}
+%  \begin{theorem}
+%    En utjämning av sannolikheterna ökar \(H(\stoch X)\).
+%  \end{theorem}
+%\end{frame}
+%
+%\begin{frame}
+%  \begin{proof}
+%  \end{proof}
+%\end{frame}
+
+\subsection{Conditional entropy}
+
+\begin{frame}
+  \begin{definition}[Conditional entropy]
+    \begin{itemize}
+      \item Define \emph{conditional entropy} \(H(\stoch Y\mid \stoch X)\) 
+        as
+        \begin{align*}
+          H(\stoch Y\mid \stoch X) = %-\sum_{i,j} \Pr(i,j)\log \Pr(j\mid i)
+          -\sum_y\sum_x \Pr(\stoch Y = y)\Pr(\stoch X = x\mid y)\log \Pr(\stoch 
+          X = x\mid y).
+        \end{align*}
+    \end{itemize}
+  \end{definition}
+
+  \pause{}
+
+  \begin{remark}
+    This is the uncertainty in \(\stoch Y\) which is not revealed by \(\stoch 
+      X\).
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{theorem}
+    \(H(\stoch X, \stoch Y) = {\color{red}H(\stoch X)} + {\color{green}H(\stoch 
+        Y\mid \stoch X)}\)
+  \end{theorem}
+
+  \begin{center}
+  \begin{tikzpicture}
+    \def\HX{(0,0) circle (2)}
+    \def\HY{(2,0) circle (2)}
+    \def\HYX{(1,0)}
+    \begin{scope}%[fill opacity=0.75]
+      \fill[green] \HY;
+      \fill[red] \HX;
+      \draw \HX node[left] {$H(X)$};
+      \draw \HY node[right] {$H(Y\mid X)$};
+    \end{scope}
+  \end{tikzpicture}
+  \end{center}
+\end{frame}
+
+%\begin{frame}
+%  \begin{proof}
+%  \end{proof}
+%\end{frame}
+
+\begin{frame}
+  \begin{corollary}
+    \(H(\stoch X\mid \stoch Y) \leq H(\stoch X)\).
+  \end{corollary}
+
+  \begin{corollary}
+    \(H(\stoch X\mid \stoch Y) = H(\stoch X)\) iff \(\stoch X\) and \(\stoch 
+      Y\) independent.
+  \end{corollary}
+\end{frame}
+%
+%\begin{frame}
+%  \begin{proof}
+%  \end{proof}
+%\end{frame}
+%
+%\begin{frame}
+%  \begin{theorem}
+%    Entropin för en Markovprocess
+%  \end{theorem}
+%\end{frame}
+
+\subsection{Information density and redundancy}
+
+\begin{frame}
+  \begin{definition}
+    \begin{itemize}
+      \item Natural language \(L\).
+      \item Stochastic variable \(\stoch P^n_L\) of strings of length \(n\).
+      \item (Alphabet \(P_L\).)
+      \item Entropy of \(L\) defined as
+        \begin{align*}
+          H_L = \lim_{n\to \infty}\frac{H(\stoch P^n_L)}{n}.
+        \end{align*}
+      \item Redundancy in \(L\) is
+        \begin{align*}
+          R_L = 1 - \frac{H_L}{\log |P_L|}.
+        \end{align*}
+    \end{itemize}
+  \end{definition}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    Meaning we have \(H_L\) bits per character in \(L\).
+  \end{remark}
+
+  \begin{example}[\cite{Shannon1948amt}]
+    \begin{itemize}
+      \item Entropy of 1--1.5 bits per character in English.
+      \item Redundancy of approximately \(1 - \frac{1.25}{\log 26} \approx 
+          0.73\).
+    \end{itemize}
+  \end{example}
+
+\end{frame}
+
+\begin{frame}
+  \begin{example}[\cite{Shannon1948amt}]
+    Two-dimensional cross-word puzzles requires redundancy of approximately 
+    \(0.5\).
+  \end{example}
+
+  \begin{example}
+    \begin{itemize}
+      \item Redundancy of \enquote{SMS languages} is lower than for 
+        \enquote{non-SMS languages}.
+
+      \item Compare \enquote{också} and \enquote{oxå}.
+
+    \end{itemize}
+  \end{example}
+
+  \begin{remark}
+    \begin{itemize}
+      \item Lower redundancy is more space-efficient.
+      \item Incurs more errors.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+%\begin{frame}
+%  \begin{itemize}
+%    \item Detta säger också att vi kan uppskatta entropin för en given 
+%      Markovprocess.
+%
+%    \item Shannon modellerade språket som en Markovprocess i sin artikel 
+%      \cite{Shannon1948amt}.
+%
+%    \item Vi kan även beräkna entropin för ett givet tillstånd i en 
+%      Markovprocess genom betingad entropi.
+%
+%  \end{itemize}
+%\end{frame}
+
+\subsection{Information gain}
+
+\begin{frame}
+  \begin{definition}
+    \begin{itemize}
+      \item Set \(U\) of possible outcomes.
+      \item Probability of outcome \(u\in U\) denoted \(p_u\).
+      \item We learn that some \emph{unknown} outcome is in \(A\subset U\).
+      \item Then the \emph{information gain} \(G(A\mid U)\) is defined as
+        \begin{align*}
+          G(A\mid U) = \log\frac{1}{\Pr(A)} = -\log\Pr(A),
+        \end{align*}
+        where \(\Pr(A) = \sum_{i\in A} p_i\).
+    \end{itemize}
+  \end{definition}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Roll of dice again]
+    \begin{itemize}
+      \item Someone rolls and we should guess the result, \(\frac{1}{6}\) 
+        chance.
+      \item We learn that it was an even number, we gain
+        \begin{align*}
+          -\log\left( \frac{1}{6} + \frac{1}{6} + \frac{1}{6}\right) =
+          -\log\frac{3}{6} = \log\frac{6}{3} = \log 2 = 1.
+        \end{align*}
+      \item The remaining uncertainty is \SI{1.58}{\bit}.
+    \end{itemize}
+  \end{example}
+
+  \pause{}
+
+  \begin{remark}
+    \begin{itemize}
+      \item \(X' = \{\fcdice{2}, \fcdice{4}, \fcdice{6}\}\)
+      \item \(H(\stoch X') = - \sum_{x\in X'} \Pr(\stoch X' = x)\log \Pr(\stoch 
+          X' = x)\)
+      \item I.e.\ \(- 3 \times \frac{1}{3}\log\frac{1}{3} \approx 1.58\).
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Dice yet again]
+    \begin{itemize}
+      \item We learn the die show less than five, i.e.\ not \fcdice{5} nor 
+        \fcdice{6}.
+      \item This yields
+        \begin{align*}
+          -\log\left( 4\times\frac{1}{6}\right) = \log\frac{6}{4}\approx 0.58
+        \end{align*}
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+
+\section[Applications]{Application in security}
+
+\subsection{Passwords}
+
+\begin{frame}
+  \begin{block}{Idea~\cite{Komanduri2011opa}}
+    \begin{itemize}
+      \item Look at different aspects of passwords individually, then 
+        summarize.
+      \item Can use \(H(x_1, x_2, \ldots, x_n) \leq H(x_1) + H(x_2) + \cdots 
+          + H(x_n)\).
+      \item This allows us to reason about bounds.
+    \end{itemize}
+  \end{block}
+\end{frame}
+
+\begin{frame}
+  \begin{example}
+    \begin{itemize}
+      \item We can look at properties such as:
+        \begin{itemize}
+          \item length,
+          \item number of and placement of character classes,
+          \item the actual characters,
+          \item \dots
+        \end{itemize}
+    \end{itemize}
+  \end{example}
+
+  \pause{}
+
+  \begin{remark}
+    \begin{itemize}
+      \item These are \emph{not independent}.
+      \item The sum will be an \emph{upper bound}.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item With an upper bound we know it's not possible to do better.
+      \item With an average we know how well most users will do.
+      \item With a lower bound we have a guarantee --- not possible!
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item If a password policy yields low entropy, it implies it's bad.
+      \item If a password policy yields high entropy, it \emph{doesn't} imply 
+        that it's good.
+    \end{itemize}
+  \end{remark}
+
+  \pause
+
+  \begin{exercise}
+    Why?
+  \end{exercise}
+\end{frame}
+
+\begin{frame}
+  \begin{figure}
+    \includegraphics[height=0.7\textheight]{password_strength.png}
+    \caption{xkcd's strip on password strength.
+    Picture: xkcd~\cite{xkcd936}.}
+  \end{figure}
+\end{frame}
+
+%\begin{frame}{En förklaring av xkcd}
+%  \begin{itemize}
+%    \item Vi har 1 miljon engelska ord: ger \(\log 10^6 \approx 20\) bitar 
+%      entropi.
+%      (xkcd använder 16 bitar, vilket ger ca 70\,000 ord, alla ord i engelskan 
+%      är inte vanliga.)
+%
+%    \item Vi kan ha inledande versal: ger 1 bit entropi.
+%
+%    \item Vi har några vanliga substitutioner: uppskattningsvis 10 stycken, 
+%      d.v.s.~3 bitar entropi.
+%
+%    \item Vi har specialtecken (ej substitution): uppskattningsvis 4 bitar 
+%      entropi.
+%
+%    \item Vi har siffror: \(\log 10\approx 3\).
+%
+%    \item Ordningen på specialtecknet och siffran: ger 1 bit entropi.
+%
+%    \item Totalt 32 bitar entropi:
+%      \begin{itemize}
+%        \item Tar minst 50 dagar med 1\,000 gissningar per sekund.
+%        \item Tar strax över en timme med 1\,000\,000 gissningar per sekund.
+%      \end{itemize}
+%
+%  \end{itemize}
+%\end{frame}
+
+\begin{frame}
+  \begin{example}[Standard password]
+    \begin{itemize}
+      \item We have
+        \begin{itemize}
+          \item 26 alphabetic characters,
+          \item 10 numbers,
+          \item 10 special characters (approximately).
+        \end{itemize}
+
+      \item This yields \(\log( 2\times 26 + 10 + 10 ) = \log 72 \approx 
+          \SI{6}{\bit}\) per password character.
+
+      \item A 10-character \emph{uniformly randomly} generated password 
+        contains \SI{60}{\bit}.
+    \end{itemize}
+  \end{example}
+
+  \pause{}
+
+  \begin{remark}
+    What happens when we require two upper and two lower-case characters, two 
+    numbers must be included?
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Four-word passphrase]
+    \begin{itemize}
+      \item We have 125\,000 words in the standard Swedish dictionary.
+      \item This yields \(\log 125\,000\approx \SI{17}{\bit}\) per word.
+      \item A four-word \emph{uniformly randomly} generated passphrase contains
+        \SI{68}{\bit}.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Random sentence]
+    \begin{itemize}
+      \item We estimated the entropy per character in a language.
+      \item It was approximately \(\SI{1.25}{\bit}\) for English.
+      \item A 20-character \emph{uniformly randomly} generated sentence would 
+        yield \SI{25}{\bit}.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item All these require uniform randomness.
+      \item Humans are bad at remembering random things.
+      \item Thus they will choose non-randomly.
+      \item The entropy will thus be (possibly much) lower.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\subsection{Research about human chosen passwords}
+
+\begin{frame}
+  \begin{example}[\citetitle{Bonneau2012lpo}~\cite{Bonneau2012lpo}]
+    \begin{itemize}
+      \item Investigates how linguistics affect the choice of multi-word 
+        passphrases.
+
+      \item Users don't choose them randomly, prefer adapted to natural 
+        language.
+
+      \item \enquote{correct horse battery staple} is preferred to 
+        \enquote{horse correct battery staple} since the first is more 
+        grammatically correct.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[\citetitle{Kuo2006hso}~\cite{Kuo2006hso}]
+    \begin{itemize}
+      \item Studied how users creates easy-to-remember passwords.
+
+      \item Also investigated the strength of phrase-based passwords.
+
+      \item E.g.\ Google's example \enquote{To be or not to be, that is the 
+          question}\footnote{%
+          URL\@: 
+          \protect\url{http://www.lightbluetouchpaper.org/2011/11/08/want-to-create-a-really-strong-password-dont-ask-google/}.
+        } which results in \enquote{2bon2btitq}.
+
+      \item This particular password has apparently been used by many \dots
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item There is a PhD thesis on the topic of guessing passwords: 
+        \fullcite{GuessingHumanChosenSecrets}.
+      \item There is even a conference dedicated to passwords: PasswordsCon.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\subsection{Identifying information}
+
+\begin{frame}
+  \begin{example}
+    Do we get more information from zodiac signs or birthdays?
+    \begin{align*}
+      -\sum_{\mathclap{\text{zodiacs}}} \frac{1}{12} \log\frac{1}{12} &= \log 12 
+      \approx 3.58 \\
+      &< -\sum_{\mathclap{\text{days of year}}} \frac{1}{365} \log\frac{1}{365} 
+      = \log 365 \approx 8.51.
+    \end{align*}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{exercise}
+    How much information do we need to uniquely identify an individual?
+  \end{exercise}
+\end{frame}
+
+\begin{frame}
+  \begin{example}
+    \begin{itemize}
+      \item Sometime during 2011 there were \(n = 6\,973\,738\,433\)\footnote{%
+          According to the World Bank.
+        } people on earth.
+
+      \item To give everyone a unique identifier we need \(\log n\approx 
+          32.7\approx 33\) bits of information.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{block}{Identifying information in browsers}
+    \begin{itemize}
+      \item Electronic Frontier Foundation (EFF) studied~\cite{Eckersley2010hui} 
+        how much information a web-browser shares.
+
+      \item You can try your browser in \url{http://panopticlick.eff.org/}.
+    \end{itemize}
+  \end{block}
+
+  \pause{}
+
+  \begin{example}[My browser]
+    \begin{itemize}
+      \item My Firefox-browser with all addons gave 21.45 bits of entropy.
+
+      \item Then the number of tested users were 2\,860\,696.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{figure}
+    \includegraphics[height=0.7\textheight]{collusion.png}
+    \caption{Screenshot from Collusion (now Lightbeam) for Firefox.
+      Map over all pages that track me using this information.}
+  \end{figure}
+\end{frame}
+
+
+%%%%%%%%%%%%%%%%%%%%%%
+
+\subsection*{References}
+\begin{frame}[allowframebreaks]
+	\small
+  \printbibliography{}
+\end{frame}
+
diff --git a/basics/preamble.tex b/basics/preamble.tex
new file mode 100644
index 0000000..b4650a3
--- /dev/null
+++ b/basics/preamble.tex
@@ -0,0 +1,23 @@
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage[swedish,british]{babel}
+\usepackage{url}
+\usepackage{graphicx}
+\usepackage{color}
+\usepackage{subfig}
+\usepackage{multicol}
+\usepackage{amssymb,amsmath,amsthm}
+\usepackage{booktabs}
+%\usepackage[squaren,binary]{SIunits}
+\usepackage[binary-units]{siunitx}
+\usepackage[strict]{csquotes}
+\usepackage{cleveref}
+\usepackage{hhcount}
+\usepackage{pgfplots}
+
+\usepackage{mathtools}
+
+\setbeamertemplate{bibliography item}[text]
+\usepackage[natbib,style=alphabetic,maxbibnames=99]{biblatex}
+\addbibresource{basics.bib}
+

From 54c78d80eef02437b1c32c65d7c17d030773645a Mon Sep 17 00:00:00 2001
From: Daniel Bosk <dbosk@kth.se>
Date: Tue, 5 Feb 2019 13:09:53 +0100
Subject: [PATCH 2/6] Rearranges sections in basics

---
 basics/contents.tex | 122 ++++++++++++++++++++++----------------------
 1 file changed, 60 insertions(+), 62 deletions(-)

diff --git a/basics/contents.tex b/basics/contents.tex
index 7f722e6..1604f73 100644
--- a/basics/contents.tex
+++ b/basics/contents.tex
@@ -9,7 +9,7 @@
 %   15 and 30 frames, all told.
 
 
-\section{Introduction}
+\section{Shannon entropy}
 
 \subsection{History}
 
@@ -55,8 +55,6 @@ \subsection{History}
 \end{frame}
 
 
-\section{Shannon entropy}
-
 \subsection{Definition of Shannon Entropy}
 
 \begin{frame}
@@ -446,6 +444,65 @@ \subsection{Conditional entropy}
 %  \end{theorem}
 %\end{frame}
 
+\subsection{Information gain}
+
+\begin{frame}
+  \begin{definition}
+    \begin{itemize}
+      \item Set \(U\) of possible outcomes.
+      \item Probability of outcome \(u\in U\) denoted \(p_u\).
+      \item We learn that some \emph{unknown} outcome is in \(A\subset U\).
+      \item Then the \emph{information gain} \(G(A\mid U)\) is defined as
+        \begin{align*}
+          G(A\mid U) = \log\frac{1}{\Pr(A)} = -\log\Pr(A),
+        \end{align*}
+        where \(\Pr(A) = \sum_{i\in A} p_i\).
+    \end{itemize}
+  \end{definition}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Roll of dice again]
+    \begin{itemize}
+      \item Someone rolls and we should guess the result, \(\frac{1}{6}\) 
+        chance.
+      \item We learn that it was an even number, we gain
+        \begin{align*}
+          -\log\left( \frac{1}{6} + \frac{1}{6} + \frac{1}{6}\right) =
+          -\log\frac{3}{6} = \log\frac{6}{3} = \log 2 = 1.
+        \end{align*}
+      \item The remaining uncertainty is \SI{1.58}{\bit}.
+    \end{itemize}
+  \end{example}
+
+  \pause{}
+
+  \begin{remark}
+    \begin{itemize}
+      \item \(X' = \{\fcdice{2}, \fcdice{4}, \fcdice{6}\}\)
+      \item \(H(\stoch X') = - \sum_{x\in X'} \Pr(\stoch X' = x)\log \Pr(\stoch 
+          X' = x)\)
+      \item I.e.\ \(- 3 \times \frac{1}{3}\log\frac{1}{3} \approx 1.58\).
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Dice yet again]
+    \begin{itemize}
+      \item We learn the die show less than five, i.e.\ not \fcdice{5} nor 
+        \fcdice{6}.
+      \item This yields
+        \begin{align*}
+          -\log\left( 4\times\frac{1}{6}\right) = \log\frac{6}{4}\approx 0.58
+        \end{align*}
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+
+\section[Applications]{Applications}
+
 \subsection{Information density and redundancy}
 
 \begin{frame}
@@ -519,65 +576,6 @@ \subsection{Information density and redundancy}
 %  \end{itemize}
 %\end{frame}
 
-\subsection{Information gain}
-
-\begin{frame}
-  \begin{definition}
-    \begin{itemize}
-      \item Set \(U\) of possible outcomes.
-      \item Probability of outcome \(u\in U\) denoted \(p_u\).
-      \item We learn that some \emph{unknown} outcome is in \(A\subset U\).
-      \item Then the \emph{information gain} \(G(A\mid U)\) is defined as
-        \begin{align*}
-          G(A\mid U) = \log\frac{1}{\Pr(A)} = -\log\Pr(A),
-        \end{align*}
-        where \(\Pr(A) = \sum_{i\in A} p_i\).
-    \end{itemize}
-  \end{definition}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Roll of dice again]
-    \begin{itemize}
-      \item Someone rolls and we should guess the result, \(\frac{1}{6}\) 
-        chance.
-      \item We learn that it was an even number, we gain
-        \begin{align*}
-          -\log\left( \frac{1}{6} + \frac{1}{6} + \frac{1}{6}\right) =
-          -\log\frac{3}{6} = \log\frac{6}{3} = \log 2 = 1.
-        \end{align*}
-      \item The remaining uncertainty is \SI{1.58}{\bit}.
-    \end{itemize}
-  \end{example}
-
-  \pause{}
-
-  \begin{remark}
-    \begin{itemize}
-      \item \(X' = \{\fcdice{2}, \fcdice{4}, \fcdice{6}\}\)
-      \item \(H(\stoch X') = - \sum_{x\in X'} \Pr(\stoch X' = x)\log \Pr(\stoch 
-          X' = x)\)
-      \item I.e.\ \(- 3 \times \frac{1}{3}\log\frac{1}{3} \approx 1.58\).
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Dice yet again]
-    \begin{itemize}
-      \item We learn the die show less than five, i.e.\ not \fcdice{5} nor 
-        \fcdice{6}.
-      \item This yields
-        \begin{align*}
-          -\log\left( 4\times\frac{1}{6}\right) = \log\frac{6}{4}\approx 0.58
-        \end{align*}
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-
-\section[Applications]{Application in security}
-
 \subsection{Passwords}
 
 \begin{frame}

From 1281ac52058a542164403792a5d3ff238bb42b16 Mon Sep 17 00:00:00 2001
From: Daniel Bosk <dbosk@kth.se>
Date: Tue, 5 Feb 2019 13:49:27 +0100
Subject: [PATCH 3/6] Breaks basics into shannon-entropy and applications

---
 Makefile                                      |   3 +-
 applications/.gitignore                       |   6 +
 {basics => applications}/Makefile             |  13 +-
 {basics => applications}/abstract.tex         |   0
 .../applications.bib                          |   0
 .../applications.tex                          |   0
 {basics => applications}/collusion.png        | Bin
 applications/contents.tex                     | 361 ++++++++++++++++++
 {basics => applications}/preamble.tex         |   2 +-
 basics/.gitignore                             |   6 -
 shannon-entropy/.gitignore                    |   6 +
 shannon-entropy/Makefile                      |  34 ++
 shannon-entropy/abstract.tex                  |  13 +
 shannon-entropy/collusion.png                 | Bin 0 -> 109951 bytes
 {basics => shannon-entropy}/contents.tex      | 342 -----------------
 shannon-entropy/preamble.tex                  |  23 ++
 shannon-entropy/shannon-entropy.bib           | 123 ++++++
 shannon-entropy/shannon-entropy.tex           |  80 ++++
 18 files changed, 654 insertions(+), 358 deletions(-)
 create mode 100644 applications/.gitignore
 rename {basics => applications}/Makefile (73%)
 rename {basics => applications}/abstract.tex (100%)
 rename basics/basics.bib => applications/applications.bib (100%)
 rename basics/basics.tex => applications/applications.tex (100%)
 rename {basics => applications}/collusion.png (100%)
 create mode 100644 applications/contents.tex
 rename {basics => applications}/preamble.tex (94%)
 delete mode 100644 basics/.gitignore
 create mode 100644 shannon-entropy/.gitignore
 create mode 100644 shannon-entropy/Makefile
 create mode 100644 shannon-entropy/abstract.tex
 create mode 100644 shannon-entropy/collusion.png
 rename {basics => shannon-entropy}/contents.tex (60%)
 create mode 100644 shannon-entropy/preamble.tex
 create mode 100644 shannon-entropy/shannon-entropy.bib
 create mode 100644 shannon-entropy/shannon-entropy.tex

diff --git a/Makefile b/Makefile
index 55b4d89..99facbd 100644
--- a/Makefile
+++ b/Makefile
@@ -1,5 +1,6 @@
-SUBDIR+= 	basics
 SUBDIR+= 	probability
+SUBDIR+= 	shannon-entropy
+SUBDIR+= 	applications
 
 INCLUDE_MAKEFILES=makefiles
 include ${INCLUDE_MAKEFILES}/subdir.mk
diff --git a/applications/.gitignore b/applications/.gitignore
new file mode 100644
index 0000000..116ecc4
--- /dev/null
+++ b/applications/.gitignore
@@ -0,0 +1,6 @@
+applications.log.gnuplot
+applications.log.table
+applications.pdf
+encryptic.png
+password_reuse.png
+password_strength.png
diff --git a/basics/Makefile b/applications/Makefile
similarity index 73%
rename from basics/Makefile
rename to applications/Makefile
index 8a9f84b..310affa 100644
--- a/basics/Makefile
+++ b/applications/Makefile
@@ -1,16 +1,13 @@
 .PHONY: all
-all: basics.pdf
+all: applications.pdf
 
-LATEXFLAGS+= 	-shell-escape
-TEX_OUTDIR= 	.
-
-SRC+=		basics.tex preamble.tex basics.bib
+SRC+=		applications.tex preamble.tex applications.bib
 SRC+= 		abstract.tex contents.tex
 SRC+=		hhcount.sty
 FIGURES+= 	collusion.png
 FIGURES+= 	password_strength.png password_reuse.png encryptic.png
 
-basics.pdf: ${SRC} ${FIGURES}
+applications.pdf: ${SRC} ${FIGURES}
 
 password_strength.png:
 	wget -O $@ http://imgs.xkcd.com/comics/password_strength.png
@@ -29,8 +26,8 @@ hhcount.sty hhutils0.sty:
 
 .PHONY: clean
 clean:
-	${RM} basics.pdf
-	${RM} basics.bbl basics.log.gnuplot basics.log.table basics.nav basics.run.xml basics.snm basics.vrb
+	${RM} applications.pdf
+	${RM} applications.bbl applications.log.gnuplot applications.log.table applications.nav applications.run.xml applications.snm applications.vrb
 
 
 .PHONY: distclean
diff --git a/basics/abstract.tex b/applications/abstract.tex
similarity index 100%
rename from basics/abstract.tex
rename to applications/abstract.tex
diff --git a/basics/basics.bib b/applications/applications.bib
similarity index 100%
rename from basics/basics.bib
rename to applications/applications.bib
diff --git a/basics/basics.tex b/applications/applications.tex
similarity index 100%
rename from basics/basics.tex
rename to applications/applications.tex
diff --git a/basics/collusion.png b/applications/collusion.png
similarity index 100%
rename from basics/collusion.png
rename to applications/collusion.png
diff --git a/applications/contents.tex b/applications/contents.tex
new file mode 100644
index 0000000..c338ddf
--- /dev/null
+++ b/applications/contents.tex
@@ -0,0 +1,361 @@
+% Since this a solution template for a generic talk, very little can
+% be said about how it should be structured. However, the talk length
+% of between 15min and 45min and the theme suggest that you stick to
+% the following rules:  
+
+% - Exactly two or three sections (other than the summary).
+% - At *most* three subsections per section.
+% - Talk about 30s to 2min per frame. So there should be between about
+%   15 and 30 frames, all told.
+
+
+\section{Applications}
+
+\subsection{Information density and redundancy}
+
+\begin{frame}
+  \begin{definition}
+    \begin{itemize}
+      \item Natural language \(L\).
+      \item Stochastic variable \(\stoch P^n_L\) of strings of length \(n\).
+      \item (Alphabet \(P_L\).)
+      \item Entropy of \(L\) defined as
+        \begin{align*}
+          H_L = \lim_{n\to \infty}\frac{H(\stoch P^n_L)}{n}.
+        \end{align*}
+      \item Redundancy in \(L\) is
+        \begin{align*}
+          R_L = 1 - \frac{H_L}{\log |P_L|}.
+        \end{align*}
+    \end{itemize}
+  \end{definition}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    Meaning we have \(H_L\) bits per character in \(L\).
+  \end{remark}
+
+  \begin{example}[\cite{Shannon1948amt}]
+    \begin{itemize}
+      \item Entropy of 1--1.5 bits per character in English.
+      \item Redundancy of approximately \(1 - \frac{1.25}{\log 26} \approx 
+          0.73\).
+    \end{itemize}
+  \end{example}
+
+\end{frame}
+
+\begin{frame}
+  \begin{example}[\cite{Shannon1948amt}]
+    Two-dimensional cross-word puzzles requires redundancy of approximately 
+    \(0.5\).
+  \end{example}
+
+  \begin{example}
+    \begin{itemize}
+      \item Redundancy of \enquote{SMS languages} is lower than for 
+        \enquote{non-SMS languages}.
+
+      \item Compare \enquote{också} and \enquote{oxå}.
+
+    \end{itemize}
+  \end{example}
+
+  \begin{remark}
+    \begin{itemize}
+      \item Lower redundancy is more space-efficient.
+      \item Incurs more errors.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+%\begin{frame}
+%  \begin{itemize}
+%    \item Detta säger också att vi kan uppskatta entropin för en given 
+%      Markovprocess.
+%
+%    \item Shannon modellerade språket som en Markovprocess i sin artikel 
+%      \cite{Shannon1948amt}.
+%
+%    \item Vi kan även beräkna entropin för ett givet tillstånd i en 
+%      Markovprocess genom betingad entropi.
+%
+%  \end{itemize}
+%\end{frame}
+
+\subsection{Passwords}
+
+\begin{frame}
+  \begin{block}{Idea~\cite{Komanduri2011opa}}
+    \begin{itemize}
+      \item Look at different aspects of passwords individually, then 
+        summarize.
+      \item Can use \(H(x_1, x_2, \ldots, x_n) \leq H(x_1) + H(x_2) + \cdots 
+          + H(x_n)\).
+      \item This allows us to reason about bounds.
+    \end{itemize}
+  \end{block}
+\end{frame}
+
+\begin{frame}
+  \begin{example}
+    \begin{itemize}
+      \item We can look at properties such as:
+        \begin{itemize}
+          \item length,
+          \item number of and placement of character classes,
+          \item the actual characters,
+          \item \dots
+        \end{itemize}
+    \end{itemize}
+  \end{example}
+
+  \pause{}
+
+  \begin{remark}
+    \begin{itemize}
+      \item These are \emph{not independent}.
+      \item The sum will be an \emph{upper bound}.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item With an upper bound we know it's not possible to do better.
+      \item With an average we know how well most users will do.
+      \item With a lower bound we have a guarantee --- not possible!
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item If a password policy yields low entropy, it implies it's bad.
+      \item If a password policy yields high entropy, it \emph{doesn't} imply 
+        that it's good.
+    \end{itemize}
+  \end{remark}
+
+  \pause
+
+  \begin{exercise}
+    Why?
+  \end{exercise}
+\end{frame}
+
+\begin{frame}
+  \begin{figure}
+    \includegraphics[height=0.7\textheight]{password_strength.png}
+    \caption{xkcd's strip on password strength.
+    Picture: xkcd~\cite{xkcd936}.}
+  \end{figure}
+\end{frame}
+
+%\begin{frame}{En förklaring av xkcd}
+%  \begin{itemize}
+%    \item Vi har 1 miljon engelska ord: ger \(\log 10^6 \approx 20\) bitar 
+%      entropi.
+%      (xkcd använder 16 bitar, vilket ger ca 70\,000 ord, alla ord i engelskan 
+%      är inte vanliga.)
+%
+%    \item Vi kan ha inledande versal: ger 1 bit entropi.
+%
+%    \item Vi har några vanliga substitutioner: uppskattningsvis 10 stycken, 
+%      d.v.s.~3 bitar entropi.
+%
+%    \item Vi har specialtecken (ej substitution): uppskattningsvis 4 bitar 
+%      entropi.
+%
+%    \item Vi har siffror: \(\log 10\approx 3\).
+%
+%    \item Ordningen på specialtecknet och siffran: ger 1 bit entropi.
+%
+%    \item Totalt 32 bitar entropi:
+%      \begin{itemize}
+%        \item Tar minst 50 dagar med 1\,000 gissningar per sekund.
+%        \item Tar strax över en timme med 1\,000\,000 gissningar per sekund.
+%      \end{itemize}
+%
+%  \end{itemize}
+%\end{frame}
+
+\begin{frame}
+  \begin{example}[Standard password]
+    \begin{itemize}
+      \item We have
+        \begin{itemize}
+          \item 26 alphabetic characters,
+          \item 10 numbers,
+          \item 10 special characters (approximately).
+        \end{itemize}
+
+      \item This yields \(\log( 2\times 26 + 10 + 10 ) = \log 72 \approx 
+          \SI{6}{\bit}\) per password character.
+
+      \item A 10-character \emph{uniformly randomly} generated password 
+        contains \SI{60}{\bit}.
+    \end{itemize}
+  \end{example}
+
+  \pause{}
+
+  \begin{remark}
+    What happens when we require two upper and two lower-case characters, two 
+    numbers must be included?
+  \end{remark}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Four-word passphrase]
+    \begin{itemize}
+      \item We have 125\,000 words in the standard Swedish dictionary.
+      \item This yields \(\log 125\,000\approx \SI{17}{\bit}\) per word.
+      \item A four-word \emph{uniformly randomly} generated passphrase contains
+        \SI{68}{\bit}.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[Random sentence]
+    \begin{itemize}
+      \item We estimated the entropy per character in a language.
+      \item It was approximately \(\SI{1.25}{\bit}\) for English.
+      \item A 20-character \emph{uniformly randomly} generated sentence would 
+        yield \SI{25}{\bit}.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item All these require uniform randomness.
+      \item Humans are bad at remembering random things.
+      \item Thus they will choose non-randomly.
+      \item The entropy will thus be (possibly much) lower.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\subsection{Research about human chosen passwords}
+
+\begin{frame}
+  \begin{example}[\citetitle{Bonneau2012lpo}~\cite{Bonneau2012lpo}]
+    \begin{itemize}
+      \item Investigates how linguistics affect the choice of multi-word 
+        passphrases.
+
+      \item Users don't choose them randomly, prefer adapted to natural 
+        language.
+
+      \item \enquote{correct horse battery staple} is preferred to 
+        \enquote{horse correct battery staple} since the first is more 
+        grammatically correct.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{example}[\citetitle{Kuo2006hso}~\cite{Kuo2006hso}]
+    \begin{itemize}
+      \item Studied how users creates easy-to-remember passwords.
+
+      \item Also investigated the strength of phrase-based passwords.
+
+      \item E.g.\ Google's example \enquote{To be or not to be, that is the 
+          question}\footnote{%
+          URL\@: 
+          \protect\url{http://www.lightbluetouchpaper.org/2011/11/08/want-to-create-a-really-strong-password-dont-ask-google/}.
+        } which results in \enquote{2bon2btitq}.
+
+      \item This particular password has apparently been used by many \dots
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{remark}
+    \begin{itemize}
+      \item There is a PhD thesis on the topic of guessing passwords: 
+        \fullcite{GuessingHumanChosenSecrets}.
+      \item There is even a conference dedicated to passwords: PasswordsCon.
+    \end{itemize}
+  \end{remark}
+\end{frame}
+
+\subsection{Identifying information}
+
+\begin{frame}
+  \begin{example}
+    Do we get more information from zodiac signs or birthdays?
+    \begin{align*}
+      -\sum_{\mathclap{\text{zodiacs}}} \frac{1}{12} \log\frac{1}{12} &= \log 12 
+      \approx 3.58 \\
+      &< -\sum_{\mathclap{\text{days of year}}} \frac{1}{365} \log\frac{1}{365} 
+      = \log 365 \approx 8.51.
+    \end{align*}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{exercise}
+    How much information do we need to uniquely identify an individual?
+  \end{exercise}
+\end{frame}
+
+\begin{frame}
+  \begin{example}
+    \begin{itemize}
+      \item Sometime during 2011 there were \(n = 6\,973\,738\,433\)\footnote{%
+          According to the World Bank.
+        } people on earth.
+
+      \item To give everyone a unique identifier we need \(\log n\approx 
+          32.7\approx 33\) bits of information.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{block}{Identifying information in browsers}
+    \begin{itemize}
+      \item Electronic Frontier Foundation (EFF) studied~\cite{Eckersley2010hui} 
+        how much information a web-browser shares.
+
+      \item You can try your browser in \url{http://panopticlick.eff.org/}.
+    \end{itemize}
+  \end{block}
+
+  \pause{}
+
+  \begin{example}[My browser]
+    \begin{itemize}
+      \item My Firefox-browser with all addons gave 21.45 bits of entropy.
+
+      \item Then the number of tested users were 2\,860\,696.
+    \end{itemize}
+  \end{example}
+\end{frame}
+
+\begin{frame}
+  \begin{figure}
+    \includegraphics[height=0.7\textheight]{collusion.png}
+    \caption{Screenshot from Collusion (now Lightbeam) for Firefox.
+      Map over all pages that track me using this information.}
+  \end{figure}
+\end{frame}
+
+
+%%%%%%%%%%%%%%%%%%%%%%
+
+\subsection*{References}
+\begin{frame}[allowframebreaks]
+	\small
+  \printbibliography{}
+\end{frame}
+
diff --git a/basics/preamble.tex b/applications/preamble.tex
similarity index 94%
rename from basics/preamble.tex
rename to applications/preamble.tex
index b4650a3..dec504b 100644
--- a/basics/preamble.tex
+++ b/applications/preamble.tex
@@ -19,5 +19,5 @@
 
 \setbeamertemplate{bibliography item}[text]
 \usepackage[natbib,style=alphabetic,maxbibnames=99]{biblatex}
-\addbibresource{basics.bib}
+\addbibresource{applications.bib}
 
diff --git a/basics/.gitignore b/basics/.gitignore
deleted file mode 100644
index f8a4f0a..0000000
--- a/basics/.gitignore
+++ /dev/null
@@ -1,6 +0,0 @@
-basics.log.gnuplot
-basics.log.table
-basics.pdf
-encryptic.png
-password_reuse.png
-password_strength.png
diff --git a/shannon-entropy/.gitignore b/shannon-entropy/.gitignore
new file mode 100644
index 0000000..e27d6d8
--- /dev/null
+++ b/shannon-entropy/.gitignore
@@ -0,0 +1,6 @@
+shannon-entropy.log.gnuplot
+shannon-entropy.log.table
+shannon-entropy.pdf
+encryptic.png
+password_reuse.png
+password_strength.png
diff --git a/shannon-entropy/Makefile b/shannon-entropy/Makefile
new file mode 100644
index 0000000..0754a58
--- /dev/null
+++ b/shannon-entropy/Makefile
@@ -0,0 +1,34 @@
+.PHONY: all
+all: shannon-entropy.pdf
+
+LATEXFLAGS+= 	-shell-escape
+TEX_OUTDIR= 	.
+
+SRC+=		shannon-entropy.tex preamble.tex shannon-entropy.bib
+SRC+= 		abstract.tex contents.tex
+SRC+=		hhcount.sty
+
+shannon-entropy.pdf: ${SRC}
+
+hhcount.sty: hhutils0.sty
+hhcount.sty hhutils0.sty:
+	wget -O $@ http://tug.ctan.org/tex-archive/usergrps/uktug/baskervi/5_5/$@
+	touch $@
+
+
+.PHONY: clean
+clean:
+	${RM} shannon-entropy.pdf
+	${RM} shannon-entropy.bbl shannon-entropy.log.gnuplot shannon-entropy.log.table shannon-entropy.nav shannon-entropy.run.xml shannon-entropy.snm shannon-entropy.vrb
+
+
+.PHONY: distclean
+distclean:
+	${RM} password_strength.png
+	${RM} password_reuse.png
+	${RM} encryptic.png
+	${RM} hhcount.sty hhutils0.sty
+
+
+INCLUDE_MAKEFILES=../makefiles
+include ${INCLUDE_MAKEFILES}/tex.mk
diff --git a/shannon-entropy/abstract.tex b/shannon-entropy/abstract.tex
new file mode 100644
index 0000000..2513055
--- /dev/null
+++ b/shannon-entropy/abstract.tex
@@ -0,0 +1,13 @@
+The area of Information Theory was founded in 1948 by Claude Shannon.
+It concerns information, e.g.\ how much information is contained in certain 
+data.
+Equivalently, it is also a measure of uncertainty in information, and has thus 
+plenty of application in security and cryptography.
+
+The concept of entropy, the main part of information theory, is treated in 
+a few short texts: \citetitle{Eckersley2010apo}~\cite{Eckersley2010apo} and 
+applied in \citetitle{Eckersley2010hui}~\cite{Eckersley2010hui}, but also in 
+\citetitle{Ueltschi2013se}~\cite{Ueltschi2013se}.
+This is then utilised in the text \citetitle{Bosk2013gl}~\cite{Bosk2013gl} (in 
+Swedish), and \citetitle{Komanduri2011opa}~\cite{Komanduri2011opa} which treats 
+passwords.
diff --git a/shannon-entropy/collusion.png b/shannon-entropy/collusion.png
new file mode 100644
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literal 0
HcmV?d00001

diff --git a/basics/contents.tex b/shannon-entropy/contents.tex
similarity index 60%
rename from basics/contents.tex
rename to shannon-entropy/contents.tex
index 1604f73..3eadb8e 100644
--- a/basics/contents.tex
+++ b/shannon-entropy/contents.tex
@@ -501,348 +501,6 @@ \subsection{Information gain}
 \end{frame}
 
 
-\section[Applications]{Applications}
-
-\subsection{Information density and redundancy}
-
-\begin{frame}
-  \begin{definition}
-    \begin{itemize}
-      \item Natural language \(L\).
-      \item Stochastic variable \(\stoch P^n_L\) of strings of length \(n\).
-      \item (Alphabet \(P_L\).)
-      \item Entropy of \(L\) defined as
-        \begin{align*}
-          H_L = \lim_{n\to \infty}\frac{H(\stoch P^n_L)}{n}.
-        \end{align*}
-      \item Redundancy in \(L\) is
-        \begin{align*}
-          R_L = 1 - \frac{H_L}{\log |P_L|}.
-        \end{align*}
-    \end{itemize}
-  \end{definition}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    Meaning we have \(H_L\) bits per character in \(L\).
-  \end{remark}
-
-  \begin{example}[\cite{Shannon1948amt}]
-    \begin{itemize}
-      \item Entropy of 1--1.5 bits per character in English.
-      \item Redundancy of approximately \(1 - \frac{1.25}{\log 26} \approx 
-          0.73\).
-    \end{itemize}
-  \end{example}
-
-\end{frame}
-
-\begin{frame}
-  \begin{example}[\cite{Shannon1948amt}]
-    Two-dimensional cross-word puzzles requires redundancy of approximately 
-    \(0.5\).
-  \end{example}
-
-  \begin{example}
-    \begin{itemize}
-      \item Redundancy of \enquote{SMS languages} is lower than for 
-        \enquote{non-SMS languages}.
-
-      \item Compare \enquote{också} and \enquote{oxå}.
-
-    \end{itemize}
-  \end{example}
-
-  \begin{remark}
-    \begin{itemize}
-      \item Lower redundancy is more space-efficient.
-      \item Incurs more errors.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-%\begin{frame}
-%  \begin{itemize}
-%    \item Detta säger också att vi kan uppskatta entropin för en given 
-%      Markovprocess.
-%
-%    \item Shannon modellerade språket som en Markovprocess i sin artikel 
-%      \cite{Shannon1948amt}.
-%
-%    \item Vi kan även beräkna entropin för ett givet tillstånd i en 
-%      Markovprocess genom betingad entropi.
-%
-%  \end{itemize}
-%\end{frame}
-
-\subsection{Passwords}
-
-\begin{frame}
-  \begin{block}{Idea~\cite{Komanduri2011opa}}
-    \begin{itemize}
-      \item Look at different aspects of passwords individually, then 
-        summarize.
-      \item Can use \(H(x_1, x_2, \ldots, x_n) \leq H(x_1) + H(x_2) + \cdots 
-          + H(x_n)\).
-      \item This allows us to reason about bounds.
-    \end{itemize}
-  \end{block}
-\end{frame}
-
-\begin{frame}
-  \begin{example}
-    \begin{itemize}
-      \item We can look at properties such as:
-        \begin{itemize}
-          \item length,
-          \item number of and placement of character classes,
-          \item the actual characters,
-          \item \dots
-        \end{itemize}
-    \end{itemize}
-  \end{example}
-
-  \pause{}
-
-  \begin{remark}
-    \begin{itemize}
-      \item These are \emph{not independent}.
-      \item The sum will be an \emph{upper bound}.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item With an upper bound we know it's not possible to do better.
-      \item With an average we know how well most users will do.
-      \item With a lower bound we have a guarantee --- not possible!
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item If a password policy yields low entropy, it implies it's bad.
-      \item If a password policy yields high entropy, it \emph{doesn't} imply 
-        that it's good.
-    \end{itemize}
-  \end{remark}
-
-  \pause
-
-  \begin{exercise}
-    Why?
-  \end{exercise}
-\end{frame}
-
-\begin{frame}
-  \begin{figure}
-    \includegraphics[height=0.7\textheight]{password_strength.png}
-    \caption{xkcd's strip on password strength.
-    Picture: xkcd~\cite{xkcd936}.}
-  \end{figure}
-\end{frame}
-
-%\begin{frame}{En förklaring av xkcd}
-%  \begin{itemize}
-%    \item Vi har 1 miljon engelska ord: ger \(\log 10^6 \approx 20\) bitar 
-%      entropi.
-%      (xkcd använder 16 bitar, vilket ger ca 70\,000 ord, alla ord i engelskan 
-%      är inte vanliga.)
-%
-%    \item Vi kan ha inledande versal: ger 1 bit entropi.
-%
-%    \item Vi har några vanliga substitutioner: uppskattningsvis 10 stycken, 
-%      d.v.s.~3 bitar entropi.
-%
-%    \item Vi har specialtecken (ej substitution): uppskattningsvis 4 bitar 
-%      entropi.
-%
-%    \item Vi har siffror: \(\log 10\approx 3\).
-%
-%    \item Ordningen på specialtecknet och siffran: ger 1 bit entropi.
-%
-%    \item Totalt 32 bitar entropi:
-%      \begin{itemize}
-%        \item Tar minst 50 dagar med 1\,000 gissningar per sekund.
-%        \item Tar strax över en timme med 1\,000\,000 gissningar per sekund.
-%      \end{itemize}
-%
-%  \end{itemize}
-%\end{frame}
-
-\begin{frame}
-  \begin{example}[Standard password]
-    \begin{itemize}
-      \item We have
-        \begin{itemize}
-          \item 26 alphabetic characters,
-          \item 10 numbers,
-          \item 10 special characters (approximately).
-        \end{itemize}
-
-      \item This yields \(\log( 2\times 26 + 10 + 10 ) = \log 72 \approx 
-          \SI{6}{\bit}\) per password character.
-
-      \item A 10-character \emph{uniformly randomly} generated password 
-        contains \SI{60}{\bit}.
-    \end{itemize}
-  \end{example}
-
-  \pause{}
-
-  \begin{remark}
-    What happens when we require two upper and two lower-case characters, two 
-    numbers must be included?
-  \end{remark}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Four-word passphrase]
-    \begin{itemize}
-      \item We have 125\,000 words in the standard Swedish dictionary.
-      \item This yields \(\log 125\,000\approx \SI{17}{\bit}\) per word.
-      \item A four-word \emph{uniformly randomly} generated passphrase contains
-        \SI{68}{\bit}.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[Random sentence]
-    \begin{itemize}
-      \item We estimated the entropy per character in a language.
-      \item It was approximately \(\SI{1.25}{\bit}\) for English.
-      \item A 20-character \emph{uniformly randomly} generated sentence would 
-        yield \SI{25}{\bit}.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item All these require uniform randomness.
-      \item Humans are bad at remembering random things.
-      \item Thus they will choose non-randomly.
-      \item The entropy will thus be (possibly much) lower.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\subsection{Research about human chosen passwords}
-
-\begin{frame}
-  \begin{example}[\citetitle{Bonneau2012lpo}~\cite{Bonneau2012lpo}]
-    \begin{itemize}
-      \item Investigates how linguistics affect the choice of multi-word 
-        passphrases.
-
-      \item Users don't choose them randomly, prefer adapted to natural 
-        language.
-
-      \item \enquote{correct horse battery staple} is preferred to 
-        \enquote{horse correct battery staple} since the first is more 
-        grammatically correct.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{example}[\citetitle{Kuo2006hso}~\cite{Kuo2006hso}]
-    \begin{itemize}
-      \item Studied how users creates easy-to-remember passwords.
-
-      \item Also investigated the strength of phrase-based passwords.
-
-      \item E.g.\ Google's example \enquote{To be or not to be, that is the 
-          question}\footnote{%
-          URL\@: 
-          \protect\url{http://www.lightbluetouchpaper.org/2011/11/08/want-to-create-a-really-strong-password-dont-ask-google/}.
-        } which results in \enquote{2bon2btitq}.
-
-      \item This particular password has apparently been used by many \dots
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{remark}
-    \begin{itemize}
-      \item There is a PhD thesis on the topic of guessing passwords: 
-        \fullcite{GuessingHumanChosenSecrets}.
-      \item There is even a conference dedicated to passwords: PasswordsCon.
-    \end{itemize}
-  \end{remark}
-\end{frame}
-
-\subsection{Identifying information}
-
-\begin{frame}
-  \begin{example}
-    Do we get more information from zodiac signs or birthdays?
-    \begin{align*}
-      -\sum_{\mathclap{\text{zodiacs}}} \frac{1}{12} \log\frac{1}{12} &= \log 12 
-      \approx 3.58 \\
-      &< -\sum_{\mathclap{\text{days of year}}} \frac{1}{365} \log\frac{1}{365} 
-      = \log 365 \approx 8.51.
-    \end{align*}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{exercise}
-    How much information do we need to uniquely identify an individual?
-  \end{exercise}
-\end{frame}
-
-\begin{frame}
-  \begin{example}
-    \begin{itemize}
-      \item Sometime during 2011 there were \(n = 6\,973\,738\,433\)\footnote{%
-          According to the World Bank.
-        } people on earth.
-
-      \item To give everyone a unique identifier we need \(\log n\approx 
-          32.7\approx 33\) bits of information.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{block}{Identifying information in browsers}
-    \begin{itemize}
-      \item Electronic Frontier Foundation (EFF) studied~\cite{Eckersley2010hui} 
-        how much information a web-browser shares.
-
-      \item You can try your browser in \url{http://panopticlick.eff.org/}.
-    \end{itemize}
-  \end{block}
-
-  \pause{}
-
-  \begin{example}[My browser]
-    \begin{itemize}
-      \item My Firefox-browser with all addons gave 21.45 bits of entropy.
-
-      \item Then the number of tested users were 2\,860\,696.
-    \end{itemize}
-  \end{example}
-\end{frame}
-
-\begin{frame}
-  \begin{figure}
-    \includegraphics[height=0.7\textheight]{collusion.png}
-    \caption{Screenshot from Collusion (now Lightbeam) for Firefox.
-      Map over all pages that track me using this information.}
-  \end{figure}
-\end{frame}
-
-
 %%%%%%%%%%%%%%%%%%%%%%
 
 \subsection*{References}
diff --git a/shannon-entropy/preamble.tex b/shannon-entropy/preamble.tex
new file mode 100644
index 0000000..ac9c38e
--- /dev/null
+++ b/shannon-entropy/preamble.tex
@@ -0,0 +1,23 @@
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage[swedish,british]{babel}
+\usepackage{url}
+\usepackage{graphicx}
+\usepackage{color}
+\usepackage{subfig}
+\usepackage{multicol}
+\usepackage{amssymb,amsmath,amsthm}
+\usepackage{booktabs}
+%\usepackage[squaren,binary]{SIunits}
+\usepackage[binary-units]{siunitx}
+\usepackage[strict]{csquotes}
+\usepackage{cleveref}
+\usepackage{hhcount}
+\usepackage{pgfplots}
+
+\usepackage{mathtools}
+
+\setbeamertemplate{bibliography item}[text]
+\usepackage[natbib,style=alphabetic,maxbibnames=99]{biblatex}
+\addbibresource{shannon-entropy.bib}
+
diff --git a/shannon-entropy/shannon-entropy.bib b/shannon-entropy/shannon-entropy.bib
new file mode 100644
index 0000000..f1efdaf
--- /dev/null
+++ b/shannon-entropy/shannon-entropy.bib
@@ -0,0 +1,123 @@
+@misc{Eckersley2010apo,
+  author={Eckersley, Peter},
+  title={A Primer on Information Theory and Privacy},
+  year={2010},
+  month={1},
+  URL={https://www.eff.org/deeplinks/2010/01/primer-information-theory-and-privacy},
+}
+@inproceedings{Eckersley2010hui,
+  author={Eckersley, Peter},
+  title={How Unique Is Your Browser?},
+  booktitle={Privacy Enhancing Technologies},
+  pages={1--18},
+  year={2010},
+  organization={Springer},
+  URL={https://panopticlick.eff.org/static/browser-uniqueness.pdf},
+}
+
+@article{Shannon1948amt,
+  author={Shannon, C. E.},
+  title={A Mathematical Theory of Communication},
+  journal={The Bell System Technical Journal},
+  volume={27},
+  pages={379--423, 623--656},
+  month={7-8},
+  year={1948},
+}
+@unpublished{Ueltschi2013se,
+  author={Ueltschi, Daniel},
+  title={Chapter 6: Shannon entropy},
+  URL={http://www.ueltschi.org/teaching/chapShannon.pdf},
+}
+
+@unpublished{Bosk2013gl,
+  author={Bosk, Daniel},
+  title={Grundl{\"a}ggande l{\"o}senordsanalys},
+  year={2013},
+  URL={http://ver.miun.se/courses/security/compendii/pwdanalysis.pdf},
+}
+
+@inproceedings{Komanduri2011opa,
+  author={Komanduri, Saranga and
+    Shay, Richard and
+    Kelley, Patrick Gage and
+    Mazurek, Michelle L. and
+    Bauer, Lujo and
+    Christin Nicolas and
+    Cranor, Lorrie Faith and
+    Egelman, Serge},
+  title={Of passwords and people:
+    Measuring the effect of password-composition policies},
+  booktitle={CHI},
+  year={2011},
+  URL={http://cups.cs.cmu.edu/rshay/pubs/passwords_and_people2011.pdf},
+}
+@inproceedings{Komanduri2014can,
+  title={Can long passwords be secure and usable?},
+  author={Shay, Richard and
+    Komanduri, Saranga and
+    Durity, Adam L and
+    Huh, Phillip Seyoung and
+    Mazurek, Michelle L and
+    Segreti, Sean M and
+    Ur, Blase and
+    Bauer, Lujo and
+    Christin, Nicolas and
+    Cranor, Lorrie Faith},
+  booktitle={Proceedings of the 32nd annual ACM conference on Human factors in 
+    computing systems},
+  pages={2927--2936},
+  year={2014},
+  organization={ACM},
+  URL={http://lorrie.cranor.org/pubs/longpass-chi2014.pdf},
+}
+
+@misc{xkcd936,
+  author={xkcd},
+  title={Password Strength},
+  URL={https://xkcd.com/936/},
+}
+@misc{xkcd792,
+  author={xkcd},
+  title={Password Reuse},
+  URL={https://xkcd.com/792/},
+}
+@misc{xkcd1286,
+  author={xkcd},
+  title={Encryptic},
+  URL={https://xkcd.com/1286/},
+}
+
+@inproceedings{Bonneau2012sog,
+  author={Bonneau, Joseph},
+  title={The science of guessing:
+    analyzing an anonymized corpus of 70 million passwords},
+  booktitle={{IEEE Symposium on Security and Privacy}},
+  year={2012},
+  URL={http://www.cl.cam.ac.uk/~jcb82/doc/B12-IEEESP-analyzing_70M_anonymized_passwords.pdf},
+}
+@inproceedings{Bonneau2012lpo,
+  author={Bonneau, Joseph and Shutova, Ekaterina},
+  title={Linguistic properties of multi-word passwords},
+  booktitle={{USEC}},
+  year={2012},
+  URL={http://www.cl.cam.ac.uk/~jcb82/doc/BS12-USEC-passphrase_linguistics.pdf},
+}
+@techreport{Kuo2006hso,
+  author={Kuo, Cynthia and Romanosky, Sasha and Cranor, Lorrie Faith},
+  title={{Human Selection of Mnemonic Phrase-based Passwords}},
+  institution={Institute of Software Research},
+  number={36},
+  year={2006},
+  URL={http://repository.cmu.edu/isr/36/},
+}
+
+@TechReport{GuessingHumanChosenSecrets,
+  author =	 {Bonneau, Joseph},
+  title = 	 {{Guessing human-chosen secrets}},
+  year = 	 2012,
+  month = 	 may,
+  url = 	 {http://www.cl.cam.ac.uk/techreports/UCAM-CL-TR-819.pdf},
+  institution =  {University of Cambridge, Computer Laboratory},
+  number = 	 {UCAM-CL-TR-819}
+}
diff --git a/shannon-entropy/shannon-entropy.tex b/shannon-entropy/shannon-entropy.tex
new file mode 100644
index 0000000..85a1f86
--- /dev/null
+++ b/shannon-entropy/shannon-entropy.tex
@@ -0,0 +1,80 @@
+%\documentclass[handout]{beamer}
+\documentclass{beamer}
+\input{preamble.tex}
+
+\usepackage{xparse}
+\ProvideDocumentEnvironment{exercise}{o}{%
+  \setbeamercolor{block body}{bg=yellow!30,fg=black}
+  \setbeamercolor{block title}{bg=yellow,fg=black}
+  \IfValueTF{#1}{%
+    \begin{block}{Exercise: #1}
+  }{%
+    \begin{block}{Exercise}
+  }
+}{%
+  \end{block}
+}
+\ProvideDocumentEnvironment{remark}{o}{%
+  \IfValueTF{#1}{%
+    \begin{alertblock}{Remark: #1}
+  }{%
+    \begin{alertblock}{Remark}
+  }
+}{%
+  \end{alertblock}
+}
+\DeclareMathOperator{\powerset}{\mathcal{P}}
+\DeclareMathOperator{\p}{\mathcal{P}}
+\let\P\p
+\DeclareMathOperator{\C}{\mathcal{C}}
+\DeclareMathOperator{\K}{\mathcal{K}}
+\DeclareMathOperator{\E}{\mathcal{E}}
+\DeclareMathOperator{\D}{\mathcal{D}}
+
+\DeclareMathOperator{\N}{\mathbb{N}}
+\DeclareMathOperator{\Z}{\mathbb{Z}}
+\DeclareMathOperator{\R}{\mathbb{R}}
+
+\let\stoch\mathbf{}
+
+\DeclareMathOperator{\xor}{\oplus}
+
+\renewcommand{\qedsymbol}{Q.E.D.}
+
+\mode<presentation>{%
+  \usetheme{Berlin}
+  \setbeamercovered{transparent}
+}
+\setbeamertemplate{footline}{\insertframenumber}
+
+\title{%
+  Applied Information Theory
+}
+\author{%
+  Daniel Bosk
+}
+\institute[MIUN IKS]{%
+  Department of Information and Communication Systems,\\
+  Mid Sweden University, Sundsvall.
+}
+\date{\today}
+
+\AtBeginSection[]{%
+  \begin{frame}
+    \tableofcontents[currentsection]
+  \end{frame}
+}
+
+\begin{document}
+
+\begin{frame}
+  \titlepage{}
+\end{frame}
+
+\begin{frame}
+  \tableofcontents
+\end{frame}
+
+\mode<all>{\input{contents.tex}}
+
+\end{document}

From 54134fcd872f012cfc5d2e1b8872202b1012393d Mon Sep 17 00:00:00 2001
From: Daniel Bosk <dbosk@kth.se>
Date: Wed, 6 Feb 2019 16:06:39 +0100
Subject: [PATCH 4/6] Adapts abstracts for shannon-entropy and applications

---
 applications/abstract.tex    | 22 +++++++++++++---------
 shannon-entropy/abstract.tex | 22 ++++++++++++++--------
 2 files changed, 27 insertions(+), 17 deletions(-)

diff --git a/applications/abstract.tex b/applications/abstract.tex
index 2513055..6bb8a86 100644
--- a/applications/abstract.tex
+++ b/applications/abstract.tex
@@ -1,13 +1,17 @@
-The area of Information Theory was founded in 1948 by Claude Shannon.
-It concerns information, e.g.\ how much information is contained in certain 
-data.
-Equivalently, it is also a measure of uncertainty in information, and has thus 
-plenty of application in security and cryptography.
+\emph{Summary:}
+There are various uses for information theory.
+In this session we will explore some of them.
 
-The concept of entropy, the main part of information theory, is treated in 
-a few short texts: \citetitle{Eckersley2010apo}~\cite{Eckersley2010apo} and 
-applied in \citetitle{Eckersley2010hui}~\cite{Eckersley2010hui}, but also in 
-\citetitle{Ueltschi2013se}~\cite{Ueltschi2013se}.
+\emph{Intended learning outcomes:}
+After this session you should be able
+\begin{itemize}
+  \item to \emph{understand} how Shannon entropy can be applied in various 
+    areas.
+\end{itemize}
+
+\emph{Reading:}
+You should read on the use of entropy to estimate anonymity:
+\citetitle{Eckersley2010hui}~\cite{Eckersley2010hui}.
 This is then utilised in the text \citetitle{Bosk2013gl}~\cite{Bosk2013gl} (in 
 Swedish), and \citetitle{Komanduri2011opa}~\cite{Komanduri2011opa} which treats 
 passwords.
diff --git a/shannon-entropy/abstract.tex b/shannon-entropy/abstract.tex
index 2513055..e3a4d1b 100644
--- a/shannon-entropy/abstract.tex
+++ b/shannon-entropy/abstract.tex
@@ -1,13 +1,19 @@
+\emph{Summary:}
 The area of Information Theory was founded in 1948 by Claude Shannon.
-It concerns information, e.g.\ how much information is contained in certain 
-data.
+It is a mathematical theory to reason about how much information is contained 
+in certain data.
 Equivalently, it is also a measure of uncertainty in information, and has thus 
 plenty of application in security and cryptography.
+This learning session covers the basic concept: Shannon entropy.
 
-The concept of entropy, the main part of information theory, is treated in 
-a few short texts: \citetitle{Eckersley2010apo}~\cite{Eckersley2010apo} and 
-applied in \citetitle{Eckersley2010hui}~\cite{Eckersley2010hui}, but also in 
+\emph{Intended learning outcomes:}
+After the session you should be able
+\begin{itemize}
+  \item to \emph{apply} Shannon entropy in basic situations.
+\end{itemize}
+
+\emph{Reading:}
+The concept of Shannon entropy, the main part of information theory, is treated 
+in a few short texts:
+\citetitle{Eckersley2010apo}~\cite{Eckersley2010apo} and
 \citetitle{Ueltschi2013se}~\cite{Ueltschi2013se}.
-This is then utilised in the text \citetitle{Bosk2013gl}~\cite{Bosk2013gl} (in 
-Swedish), and \citetitle{Komanduri2011opa}~\cite{Komanduri2011opa} which treats 
-passwords.

From 8ce8a08ac7c0b86f91efc181e421d507a187c701 Mon Sep 17 00:00:00 2001
From: Daniel Bosk <dbosk@kth.se>
Date: Wed, 6 Feb 2019 16:14:32 +0100
Subject: [PATCH 5/6] Updates slide titles

---
 applications/applications.tex       | 2 +-
 shannon-entropy/shannon-entropy.tex | 2 +-
 2 files changed, 2 insertions(+), 2 deletions(-)

diff --git a/applications/applications.tex b/applications/applications.tex
index 85a1f86..90dd7d9 100644
--- a/applications/applications.tex
+++ b/applications/applications.tex
@@ -48,7 +48,7 @@
 \setbeamertemplate{footline}{\insertframenumber}
 
 \title{%
-  Applied Information Theory
+  Applications of information theory
 }
 \author{%
   Daniel Bosk
diff --git a/shannon-entropy/shannon-entropy.tex b/shannon-entropy/shannon-entropy.tex
index 85a1f86..03e9e94 100644
--- a/shannon-entropy/shannon-entropy.tex
+++ b/shannon-entropy/shannon-entropy.tex
@@ -48,7 +48,7 @@
 \setbeamertemplate{footline}{\insertframenumber}
 
 \title{%
-  Applied Information Theory
+  Shannon entropy
 }
 \author{%
   Daniel Bosk

From d98e614fe406cef4d9ae105d562350b49a17ba6b Mon Sep 17 00:00:00 2001
From: Daniel Bosk <dbosk@kth.se>
Date: Wed, 6 Feb 2019 16:16:27 +0100
Subject: [PATCH 6/6] Adds links to recordings for shannon-entropy and
 applications

---
 applications/README.md    | 2 ++
 shannon-entropy/README.md | 2 ++
 2 files changed, 4 insertions(+)
 create mode 100644 applications/README.md
 create mode 100644 shannon-entropy/README.md

diff --git a/applications/README.md b/applications/README.md
new file mode 100644
index 0000000..2e87c69
--- /dev/null
+++ b/applications/README.md
@@ -0,0 +1,2 @@
+- "Applications of information theory" (MP4 H.264 AAC)
+  \[[HTTPS](https://ver.miun.se/courses/security/dasak/infotheory-applications.mp4)\]
diff --git a/shannon-entropy/README.md b/shannon-entropy/README.md
new file mode 100644
index 0000000..c8da595
--- /dev/null
+++ b/shannon-entropy/README.md
@@ -0,0 +1,2 @@
+- "Shannon entropy" (MP4 H.264 AAC)
+  \[[HTTPS](https://ver.miun.se/courses/security/dasak/shannon-entropy.mp4)\]