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\begin{document}
\title{An upper estimate of the amount of best solutions in labyrinths.}
\author{Patrick Hammer}
\date{September 7, 2011}
\maketitle
\section{Definitions}
\subsection{Connected Surface}
Let $x$ and $y$ be points in the surface $A$.
$A$ is a connected surface when: $\forall x,y,x\neq y:$ $\exists$ Path$(A,x,y)\subseteq A$.
\subsection{Shortest Path}
Let $A$ be a connected surface.
The shortest path $S(A,x,y)$ is one of the subsets of the shortest path set $M_S(A,x,y):=\{$Path$(A,x,y)$ $|$ Length$($Path$(A,x,y)) = $ inf$($Length$(\{$Path$(A,x,y)$ $\subseteq$ Paths$(A,x,y)\}$)$))\}$.
\subsection{Hole}
A hole $H(A)$ of a connected surface $A$, is a connected surface, which is infolded by $A$, so the maximum shortest path length between two points in $H(A)$ is limited by $A$, and there doesn't exist a point which is in both, $H(A)$ and $A$. The hole set $M_H(A)$ contains all holes of $A$.
\section{Proposition}
\subsection{Hammer's Labyrinth Proposition - Max. amount of shortest paths}
Let $A$ be a connected surface, $x,y$ points in it.
The amount of shortest paths $|M_S(A,x,y)|$ is as follows:
\begin{equation}
\nonumber $|M_S(A,x,y)| \leq 2^{|M_H(A)|}$
\end{equation}
Proof: We want to get from $x$ to $y$. Because of the triangle inequality,
we can reach a point behind a new hole, by a maximum of two best ways: By bypassing it clockwise,
and by bypassing it counterclockwise. So every new hole doubles the possibilities at most.
\subsection{Hammer's Labyrinth}
This is the sort of labyrinth which delivers equality for the previous proposition,
it has interesting symmetry properties.
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