Remove trailing spaces

The commands

find . -type f -name '*.md' -exec sed --in-place 's/[[:space:]]\+$//' {} \+

and

find . -type f -name '*.tex' -exec sed --in-place 's/[[:space:]]\+$//' {} \+

were used to do so.

documents/proof-of-correctness-pogo/proof-of-correctness-pogo.tex

@@ -14,11 +14,11 @@
 \clubpenalty  = 10000   % Schusterjungen verhindern
 \widowpenalty = 10000   % Hurenkinder verhindern
 
-\hypersetup{ 
-  pdfauthor   = {Martin Thoma}, 
-  pdfkeywords = {Google Code Jam, Round 1C 2013, Pogo}, 
-  pdftitle    = {Proof of correctness for an algorithm for pogo} 
-} 
+\hypersetup{
+  pdfauthor   = {Martin Thoma},
+  pdfkeywords = {Google Code Jam, Round 1C 2013, Pogo},
+  pdftitle    = {Proof of correctness for an algorithm for pogo}
+}
 
 % From http://www.matthewflickinger.com/blog/archives/2005/02/20/latex_mod_spacing.asp
 % Thanks!
@@ -43,7 +43,7 @@ have to find a way to get to one coordinate $(x,y) \in \mathbb{Z} \times \mathbb
 $(0, 0)$.
 
 In your
-$i$-th step you move either $\underbrace{(+i,0)}_{=: E}$, 
+$i$-th step you move either $\underbrace{(+i,0)}_{=: E}$,
 $\underbrace{(-i,0)}_{=: W}$, $\underbrace{(0,+i)}_{=: N}$ or
 $\underbrace{(0,-i)}_{=: S}$.
 
@@ -114,7 +114,7 @@ It's enough to proof $s \geq s_{\min}$ and $s \leq s_{\min}$.
 \begin{myindentpar}{1cm}
 \textbf{Theorem: } $s \leq s_{\min}$ (we don't make too many steps)
 
-\textbf{Proof: } 
+\textbf{Proof: }
 \begin{myindentpar}{1cm}
 We have to get from $(0,0)$ to $(x, y)$. As we may only move in
 taxicab geometry we have to use the taxicab distance measure $d_1$:
@@ -123,23 +123,23 @@ taxicab geometry we have to use the taxicab distance measure $d_1$:
 So in our scenario:
 \[d_1 \left ((0,0), (x,y) \right ) = |x| + |y|\]
 
-This means we have to move at least $|x| + |y|$ units to get 
+This means we have to move at least $|x| + |y|$ units to get
 from $(0,0)$ to $(x, y)$. As we move $i$ units in the $i$'th step,
 we have to solve the following equations for $s_{\min1}$:
 \begin{align}
     \sum_{i=1}^{s_{\min1}} i          &\geq |x| + |y| &&\text{ and } &|x| + |y|      &> \sum_{i=1}^{s_{\min1} - 1} i\\
-    \frac{s_{\min1}^2 + s_{\min1}}{2} &\geq |x| + |y| &&             &               &> \sum_{i=1}^{s_{\min1} - 1} i & 
+    \frac{s_{\min1}^2 + s_{\min1}}{2} &\geq |x| + |y| &&             &               &> \sum_{i=1}^{s_{\min1} - 1} i &
 \end{align}
 
-This is what algorithm \ref{alg:calculateSteps} check with \texttt{condition 1}. 
-As the algorithm increases $s$ only by one in each loop, it makes 
+This is what algorithm \ref{alg:calculateSteps} check with \texttt{condition 1}.
+As the algorithm increases $s$ only by one in each loop, it makes
 sure that $\sum_{i=1}^{s_{\min1} - 1} i$ is bigger than $|x| + |y|$.
 
 You can undo moves by going back. But this will always make an even
-number undone. When you go $(+i, 0)$ and later $(-j, 0)$ it is the 
+number undone. When you go $(+i, 0)$ and later $(-j, 0)$ it is the
 same as if you've been going $(i-j, 0)$. So $2\cdot i$ steps got undone.
 But $2\cdot i$ is an even number. You will never be able to undo
-an odd number of moved units. This means, the parity of the minimum 
+an odd number of moved units. This means, the parity of the minimum
 number of units you would have to move if you would move one unit per
 step has to be the same as the parity of the moves you actually do.
 This is exactly what \texttt{condition 2} makes sure.
@@ -149,13 +149,13 @@ So we need at least $s$ steps $\Rightarrow s \leq s_{\min} \square$
 
 \textbf{Theorem: } $s \geq s_{\min}$ (we make enough steps)
 
-\textbf{Proof: } 
+\textbf{Proof: }
 \begin{myindentpar}{1cm}
 We chose $s$ in a way that \texttt{condition 1} is true.
 As we have to go $i \in 1,\dots,s$, we can get every possible sum $\Sigma \in \Set{-\frac{s^2+s}{2}, \dots +\frac{s^2+s}{2}}$
 with a subset of $\Set{1, \dots, s}$\footnote{This can easily be proved by induction over $\Sigma$.}.
 This means we can make a partition $(A, \underbrace{\Set{1, \dots, s} \setminus A}_{=: B})$
-such that $|\sum_{i \in A} i| = |x|$ and $|\sum_{i \in B} i|-2\cdot j = |y|$. 
+such that $|\sum_{i \in A} i| = |x|$ and $|\sum_{i \in B} i|-2\cdot j = |y|$.
 This means, we can reach $(x,y)$ from $(0,0)$.
 \end{myindentpar}
 \end{myindentpar}
@@ -163,10 +163,10 @@ This means, we can reach $(x,y)$ from $(0,0)$.
 \subsection{solvePogo}
 \textbf{Theorem: } \Call{solvePogo}{$x,y$} returns a valid, minimal sequence of steps to get from $(0, 0)$ to $(x,y)$
 
-\textbf{Proof: } 
+\textbf{Proof: }
 \begin{myindentpar}{1cm}
-As $s_{\min}$ is the minimum amount of steps you need to get from 
-$(0,0)$ to $(x,y)$, \Call{solvePogo}{$x,y$} will return a minimal 
+As $s_{\min}$ is the minimum amount of steps you need to get from
+$(0,0)$ to $(x,y)$, \Call{solvePogo}{$x,y$} will return a minimal
 sequence of steps to get from $(0, 0)$ to $(x,y)$ (see proof above).
 
 We only have to prove that the sequence of steps that \Call{solvePogo}{$x,y$}