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+\documentclass[a4paper]{scrartcl}
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+\usepackage{amssymb, amsmath} % needed for math
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+\usepackage[utf8]{inputenc} % this is needed for umlauts
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+\usepackage[ngerman]{babel} % this is needed for umlauts
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+\usepackage[T1]{fontenc} % this is needed for correct output of umlauts in pdf
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+\usepackage[margin=2.5cm]{geometry} %layout
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+\usepackage{hyperref} % links im text
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+\usepackage{parskip} % no indentation on new paragraphs
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+\usepackage{color}
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+\usepackage{framed}
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+\usepackage{enumerate} % for advanced numbering of lists
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+\usepackage{algorithm,algpseudocode}
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+\usepackage{braket} % needed for \Set
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+\clubpenalty = 10000 % Schusterjungen verhindern
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+\widowpenalty = 10000 % Hurenkinder verhindern
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+
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+\hypersetup{
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+ pdfauthor = {Martin Thoma},
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+ pdfkeywords = {Google Code Jam, Round 1C 2013, Pogo},
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+ pdftitle = {Proof of correctness for an algorithm for pogo}
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+}
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+
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+% From http://www.matthewflickinger.com/blog/archives/2005/02/20/latex_mod_spacing.asp
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+% Thanks!
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+\makeatletter
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+\def\imod#1{\allowbreak\mkern10mu({\operator@font mod}\,\,#1)}
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+\makeatother
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+
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+\renewcommand{\algorithmicrequire}{\textbf{Input: }}
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+\renewcommand{\algorithmicensure}{\textbf{Output: }}
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+
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+\newenvironment{myindentpar}[1]%
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+ {\begin{list}{}%
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+ {\setlength{\leftmargin}{#1}}%
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+ \item[]%
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+ }
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+ {\end{list}}
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+
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+\begin{document}
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+\section{The Problem}
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+You're on a two-dimensional grid $\mathbb{Z} \times \mathbb{Z}$ and
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+have to find a way to get to one coordinate $(x,y) \in \mathbb{Z} \times \mathbb{Z}$. You start at
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+$(0, 0)$.
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+
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+In your
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+$i$-th step you move either $\underbrace{(+i,0)}_{=: E}$,
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+$\underbrace{(-i,0)}_{=: W}$, $\underbrace{(0,+i)}_{=: N}$ or
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+$\underbrace{(0,-i)}_{=: S}$.
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+
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+\section{The algorithm}
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+\begin{algorithm}
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+ \begin{algorithmic}
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+ \Function{calculateSteps}{$x \in \mathbb{Z}$, $y \in \mathbb{Z}$}
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+ \State $s \gets 1$
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+ \State $dist \gets |x| + |y|$
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+ \\
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+ \While{$\overbrace{\frac{s^2 + s}{2} < dist}^\text{condition 1}$ or $\overbrace{\frac{s^2 + s}{2} \not\equiv dist \imod{2}}^\text{condition 2}$}
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+ \State $s \gets s + 1$
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+ \EndWhile
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+ \\
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+ \State \Return $s$
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+ \EndFunction
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+ \end{algorithmic}
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+ \caption{Algorithm to calculate the minimum amount of steps}
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+ \label{alg:calculateSteps}
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+\end{algorithm}
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+
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+\clearpage
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+
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+\begin{algorithm}[ht!]
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+ \begin{algorithmic}[ht!]
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+ \Function{solvePogo}{$x \in \mathbb{Z}$, $y \in \mathbb{Z}$}
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+ \State $max \gets$ \Call{calculateSteps}{$x, y$}
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+ \\
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+ \State $solution \gets \varepsilon$
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+ \For{$i$ in $max, \dots, 1$}
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+ \If{$|x| > |y|$}
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+ \If{$x > 0$}
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+ \State $solution \gets solution + E$
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+ \State $x \gets x - i$
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+ \Else
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+ \State $solution \gets solution + W$
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+ \State $x \gets x + i$
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+ \EndIf
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+ \Else
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+ \If{$y > 0$}
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+ \State $solution \gets solution + N$
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+ \State $y \gets y - i$
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+ \Else
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+ \State $solution \gets solution + S$
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+ \State $y \gets y + i$
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+ \EndIf
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+ \EndIf
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+ \EndFor
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+ \\
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+ \State \Return $solution$
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+ \EndFunction
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+ \end{algorithmic}
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+ \caption{Algorithm to solve the pogo problem}
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+ \label{alg:solvePogo}
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+\end{algorithm}
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+
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+\section{Correctness}
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+\subsection{calculateSteps}
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+Let $x,y \in \mathbb{Z}$ and $s := \Call{calculateSteps}{x, y}$.
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+
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+Let $s_{\min}$ be the minimum amount of necessary steps to get from $(0,0)$
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+to $(x,y)$ when you move $i$ units in your $i$'th step.
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+
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+\textbf{Theorem: } $s = s_{\min}$
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+
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+It's enough to proof $s \geq s_{\min}$ and $s \leq s_{\min}$.
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+
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+\begin{myindentpar}{1cm}
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+\textbf{Theorem: } $s \leq s_{\min}$ (we don't make too many steps)
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+
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+\textbf{Proof: }
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+\begin{myindentpar}{1cm}
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+We have to get from $(0,0)$ to $(x, y)$. As we may only move in
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+taxicab geometry we have to use the taxicab distance measure $d_1$:
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+\[d_1 \left (p, q \right ) := \sum_{i=1}^2 |p_i -q_i|\]
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+
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+So in our scenario:
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+\[d_1 \left ((0,0), (x,y) \right ) = |x| + |y|\]
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+
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+This means we have to move at least $|x| + |y|$ units to get
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+from $(0,0)$ to $(x, y)$. As we move $i$ units in the $i$'th step,
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+we have to solve the following equations for $s_{\min1}$:
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+\begin{align}
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+ \sum_{i=1}^{s_{\min1}} i &\geq |x| + |y| &&\text{ and } &|x| + |y| &> \sum_{i=1}^{s_{\min1} - 1} i\\
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+ \frac{s_{\min1}^2 + s_{\min1}}{2} &\geq |x| + |y| && & &> \sum_{i=1}^{s_{\min1} - 1} i &
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+\end{align}
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+
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+This is what algorithm \ref{alg:calculateSteps} check with condition 1.
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+As the algorithm increases $s$ only by one in each loop, it makes
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+sure that $\sum_{i=1}^{s_{\min1} - 1} i$ is bigger than $|x| + |y|$.
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+
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+TODO: Proof necessarity of condition two
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+
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+TODO: I guess I should initialize $s$ with 0 (should only make a difference when (x,y) = (0,0))
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+\end{myindentpar}
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+
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+\textbf{Theorem: } $s \geq s_{\min}$ (we make enough steps)
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+
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+\textbf{Proof: }
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+\begin{myindentpar}{1cm}
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+TODO
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+\end{myindentpar}
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+\end{myindentpar}
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+
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+\subsection{solvePogo}
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+\textbf{Theorem: } \Call{solvePogo}{$x,y$} returns a valid, minimal sequence of steps to get from $(0, 0)$ to $(x,y)$
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+
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+\textbf{Proof: }
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+\begin{myindentpar}{1cm}
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+TODO
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+\end{myindentpar}
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+
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+\end{document}
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