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\author{Jon Doe}
\title{My Fun with Algorithms}
\date{18 April 2018}
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\section{Introduction}
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% A \label that is created for a figure, table, section etc. can later be used to refer to it using \ref (creating the number) or \pageref (creating the number of the page containing the object)
\label{sec:intro}
% With \cite, you refer to literature.
% The ~ in front of \cite{} creates a non-breaking space.
Write whatever you want to write, bablabla.
This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph. This is a new paragraph.
The \emph{Burrows-Wheeler-Transform} introduced by Burrows and Wheeler in 1994~\cite{BUR-WHE-1994} allows to, \ldots
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\section{Burrows-Wheeler-Transform}
\label{sec:bwt}
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As motivated in Section~\ref{sec:intro}, \ldots, ca.\ 1 liter of water visualized in Figure \ref{fig:example}. blabbdf jklkjf hdlkjfslkdjhflskdjflk köfjhdkjhsdlkj hlkdsj hfsdalkj fhlskdjfh lskdjjhf sldakjj hflkjs hflksdjhf lksdjh flkjsh flkj
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\begin{figure}[htb]
\begin{center}
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\caption{Example of a figure. Usually, the caption begins with a short phrase like the one before, followed by further explanations in complete sentences. If a figure is copied from another source, you have to indicate this clearly like this. Figure reprinted from~\cite{alts}.}
\label{fig:example}
\end{figure}
\subsection{Inverse Transform}
\label{sec:ibwt}
% Formulas can be embedded in the main text or separated and centered...
The inverse Burrows-Wheeler-Transform $bwt^{-1}$ can be efficiently computed as follows:
\[
bwt^{-1} := \int\limits_{0}^{\infty} x :-) .
\]
Table~\ref{tab:tabelle} shows a list of irrelevant\footnote{Avoid footnotes as they interrupt the reading flow.} numbers. And here is just some more text to produce some lines. Ignore the content---it is stupid blabla.
% A table as floating object. Similar to figures...
\begin{table}[htb]
\caption{A table. Again, further explanations follow in complete sentences and sources hav to be cited clearly.}
\label{tab:tabelle}
\begin{center}
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\hline
Aaaaa & B & C \\\hline
1 & $\pi$ & 3.01 \\
4 & forty-two & 6789.20\\
\hline
\end{tabular}
\end{center}
\end{table}
If necessary, you can also cite websites~\cite{citingWebsites,wiki:xxx2}. And again, some more text to produce some more lines. You can see that, at least in this template, a paragraph begins with an indentation.
To begin a new paragraph, there has to be at least one empty line in-between the paragraphs.
\section{Some math}
This section contains some loose collection of math to exemplify (i) some Latex syntax, and (ii) good style in presenting mathematics.
\medskip
%If you do not want to have indentation for such a short line, use \noindent in front of the line.
\noindent Let $\langle V \rangle$ be a vector space defined by
\begin{equation}\label{eqn:vs}
\langle V \rangle \ = \ \left\{ \sum\limits_{i=1}^n \alpha_i x_i \mid \alpha_i \in F, 1 \leq i \leq n \right\}.
\end{equation}
We now show that $\langle V \rangle$ is closed under addition. Consider two vectors $x,y \in \langle V \rangle$. Then $x = \sum_{i=1}^n \alpha_i x_i$ and $y = \sum_{i=1}^n \beta_i x_i$.
For any constants $\alpha, \beta \in F$, we have
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\begin{align*}
\alpha x + \beta y & \ = \ \alpha\left( \sum\nolimits_{i=1}^n \alpha_i x_i \right) + \beta\left( \sum\nolimits_{i=1}^n \beta_i x_i \right) \nonumber\\
& \ = \ \sum\limits_{i=1}^n \left( \alpha\alpha_i + \beta\beta_i \right) x_i,
\end{align*}
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so that $\alpha x + \beta y \in \langle V \rangle$, where
\[ \delta_{ij} := \left\{ \begin{array}{cl}
1 & \text{if }i=j,\\
0 & \text{otherwise.}
\end{array}
\right. \]
Based on the above observation, we now formalize the phenomenon that deleting two different intervals from a given reference results in the same string.
Let $|S|$ be the length of a string $S$.
We denote deletions by intervals $[i,j]$, where $S\backslash[i,j]$ denote the string that results from removing all characters from (and including) position $i$ to (and including) $j$ from $S$.
As introduced by Wittler \textit{et al.}~\cite{WIT-MAN-PAT-STO-2011}, deleting different segments of a given string can yield the same result.
Such deletions are called \emph{equivalent}.
\begin{definition}[\cite{WIT-MAN-PAT-STO-2011}]\label{def:equivalent}
Deletions $[i,j]$ and $[i',j']$ are \emph{equivalent} w.r.t.\ a reference sequence~$S$, written $[i,j] \Leftrightarrow [i',j']$, if and only if $S \backslash [i,j] = S \backslash [i',j']$. We further define their \emph{shift} being $s\left([i,j],[i',j']\right) := |i'-i|$.
\end{definition}
Out of a set of equivalent deletions, we are especially interested in the leftmost and rightmost representative, formally defined as follows.
\begin{definition}[rightmost and leftmost shift]
Let $S$ be a reference string. Then the \emph{leftmost} and \emph{rightmost shift} of a deletion $[i,j]$ are defined as:
$L([i,j]) := $ \linebreak[4] $\argmin\left\{\, i' \mid [i',j']\Leftrightarrow[i,j] \right\}$ and
$R([i,j]) := $ $\argmax\left\{\, i' \mid [i',j']\Leftrightarrow[i,j] \right\}$, respectively.
\end{definition}
Our main result is that deciding the mC1P is tractable for a large
family of matrices with constraints on the maximum number of entries
$1$ in multicolumns a row can have.
\begin{theorem}\label{thm:main}
Let $M$ be a binary matrix and $\mu $ a multiplicity vector such
that \ldots. Deciding if $M$ has the mC1P
for $\mu $ can be done in polynomial time and space.
\end{theorem}
\begin{proof}
We split the proof into two parts. First we consider the case where
$M$ with multiplicity vector $\mu $ contains a single multicolumn. Then we show how to handle the general
case based on Eulerian
cycles.
\ldots
\end{proof}
\medskip
You can easily refer to equations, definitions, etc.\ like in the following example. Theorem~\ref{thm:main} is neither based on Definition~\ref{def:equivalent} nor on Equation~\eqref{eqn:vs}.
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