tex fixes
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@@ -271,24 +271,25 @@
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As new states $\mStateVec_{t}$ should approach the pedestrian's destination
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As new states $\mStateVec_{t}$ should approach the pedestrian's destination
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we use a reference $\pathRef$ all states try to reach. This references must
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we use a reference $\pathRef$ all states try to reach. This references must
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be both part of the shortest path and located somewhere outside of the sample-set.
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be a) part of the shortest path and b) located somewhere outside of the sample-set.
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We thus calculate the standard deviation of the distance of all sample-positions
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For b) we need to know the current position distribution and therefore calculate
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$\fPos{\mStateVec_{t-1}}$ from aforementioned centre $\pathCentroid$.
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the standard deviation of the distance of all sample-positions $\fPos{\mStateVec_{t-1}}$
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from aforementioned centre $\pathCentroid$.
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%\commentByFrank{so klarer? platz fuer groese Eq. fehlt und Notation zum ansprechen jedes einzelnen Particles vermeide ich lieber...}
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%\commentByFrank{so klarer? platz fuer groese Eq. fehlt und Notation zum ansprechen jedes einzelnen Particles vermeide ich lieber...}
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%\begin{equation}
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%\begin{equation}
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% d_\text{cen} = \| pos(q_{t-1}) - \pathCentroid \|
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% d_\text{cen} = \| pos(q_{t-1}) - \pathCentroid \|
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% \sigma_\text{cen} = stdDev(distance)
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% \sigma_\text{cen} = stdDev(distance)
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%\end{equation}
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%\end{equation}
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After advancing the starting-vertex by three times this deviation
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After advancing the starting-vertex by three times this deviation
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we get the new point $\pathRef$ that is: part of the shortest path, outside of the sample-set
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we get the new point $\pathRef$ that is: part of the shortest path, outside of the sample-set
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and closer (but not too close) to the desired destination.
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and closer (but not too close) to the desired destination.
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Hereafter, the simple transition \refeq{eq:transSimple} is combined with a second probability,
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Hereafter, the simple transition \refeq{eq:transSimple} is combined with a second probability,
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downvoting all grid-steps that depart from $\pathRef$.
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downvoting all grid-steps that depart from $\pathRef$.
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To still allow leaving the shortest path, the intensity of the downvoting is controlled via $\mUsePath$,
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To still allow leaving the shortest path, the intensity of downvoting is controlled via $\mUsePath$,
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with $0 < \mUsePath < 1$.
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with $0 < \mUsePath < 1$.
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Finally, \refeq{eq:transShortestPath} provides a metric tending towards the reference while
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Finally, \refeq{eq:transShortestPath} provides a metric tending towards the reference while
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still allowing the pedestrian to leave the shortest path:
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still allowing the pedestrian to leave the shortest path:
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@@ -335,7 +336,9 @@
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\end{equation}
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\end{equation}
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Fig. \ref{fig:multiHeatMap} shows a heat-map of how often vertices were visited after several
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Fig. \ref{fig:multiHeatMap} shows a heat-map of how often vertices were visited after several
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\SI{125}{\meter} walks. The colours from cold to hot indicate that both possible paths
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random walks from $\mVertexA$ towards $\mVertexDest$.
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%\SI{125}{\meter} walks.
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The colours from cold to hot indicate that both possible paths
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are covered and slight deviations from the shortest version are possible.
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are covered and slight deviations from the shortest version are possible.
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\begin{figure}
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\begin{figure}
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@@ -348,7 +351,7 @@
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%\commentByFrank{so besser?}
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%\commentByFrank{so besser?}
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\label{fig:multiHeatMap}
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\label{fig:multiHeatMap}
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\end{figure}
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\end{figure}
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It should be noted that the normal distribution in \eqref{eq:transShortestPath} and \eqref{eq:transMultiPath} does not integrate to \SI{1}{} due to circularity of angular data.
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It should be noted that the normal distribution in \eqref{eq:transShortestPath} and \eqref{eq:transMultiPath} does not integrate to \SI{1}{} due to circularity of angular data.
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This normally requires wrapped distributions like the von Mises distribution.
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This normally requires wrapped distributions like the von Mises distribution.
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However, in our case, the normal distribution can be assumed as sufficient for small enough $\sigma^2$.
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However, in our case, the normal distribution can be assumed as sufficient for small enough $\sigma^2$.
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