text:
\section{Introduction}

Let $D$ be a Jordan domain, i.e.\ the homeomorphic image of a disk in the plane. Suppose that continuously-moving intruders wander in $D$. You and a friend are each given one end of a rope, and your task is to drag this rope through the domain $D$ in such a way so that every intruder is eventually intersected or caught by the rope. What is the shortest rope length you need in order to catch every possible intruder? We refer to such a continuous rope motion as a \emph{sweep} of $D$ (Figure~\ref{fig:sweep}), and we refer to the length of the shortest such possible rope as the \emph{sweeping cost} of $D$.

\begin{figure}[h]
	\def6.5in{5in}
	\import{}{intro_diagram.pdf_tex}
\caption{An example sweep of a domain in the plane. A time $t=0$ the entire domain is contaminated, and at time $t=1$ the clearing sweep is complete.}
\label{fig:sweep}
\end{figure}

The problem we consider is only one example of a wide variety of interesting pursuit-evasion problems; see Section~\ref{sec:related} for a brief introduction or \cite{chung2011search}, for example, for a survey. It is a pursuit-evasion problem in which both space and time are continuous, the pursuer is modeled at each point in time by a continuous curve, the intruder has complete information about the pursuer's location and its planned future movements, and the pursuer has no knowledge of the intruder's movements. Our problem can also be phrased as a contamination-clearing task, in which one must find the shortest rope necessary to clear domain $D$ of a contaminant which, when otherwise unrestricted by the rope, moves at infinite speed to fill its region.

As first examples, the sweeping cost of a disk is equal to its diameter, and the sweeping cost of an ellipse is equal to the length of its minor axis. These computations follow from Theorem~\ref{thm:area-lower-bound}, in which we prove that the sweeping cost of domain $D$ is at least as large as the shortest area-bisecting curve in $D$.

One motivation for considering a pursuer which is a rope, or a continuous curve at each point in time, is the context of mobile sensor networks. Suppose there is a large collection of disk-shaped sensors moving inside a planar domain, as considered in \cite{de2006coordinate,adams2015evasion}. What is the minimal number of sensors needed to clear this domain of all possible intruders? If $n$ is the number of sensors, and $\frac{1}{n}$ is the diameter of each sensor, then as $n\to\infty$ an upper bound for the number of sensors needed is given by the sweeping cost.

\begin{figure}[h]
	\def6.5in{5in}
	\import{}{robots_to_line.pdf_tex}
\label{fig:robots_to_line}
\end{figure}

As our main result, in Theorem~\ref{thm:frechet} we provide an analytic formula for the sweeping cost of an arbitrary Jordan domain $D$. Indeed, the sweeping cost of $D$ is equal to the infimum, taken over all pairs of curves in the boundary of $D$ whose concatenation wraps a nontrivial number of times around the boundary, of the geodesic Fr\'{e}chet distance between the two curves. The geodesic Fr\'{e}chet distance differs from the standard Fr\'{e}chet distance in that the distance between two points in $D$ is not their Euclidean distance, but instead the length of the shortest path between them in $D$. Our Theorem~\ref{thm:frechet} is also closely related to the geodesic width between two curves \cite{efrat2002new}.

Using our main result, we prove in Theorem~\ref{thm:sc=width} that the sweeping cost of a convex domain $D$ is equal to the width of $D$. As a consequence, it follows that the sweeping cost of a polygonal convex domain with $n$ vertices can be computed in time $O(n)$ and space $O(n)$ using the \emph{rotating calipers} technique \cite{toussaint1983solving}. Furthermore, it follows from \cite{pal1921minimumproblem,cerdan2006comparing} that a convex domain of unit area with the maximal possible sweeping cost---i.e.\ the most expensive convex domain to clear with a rope---is the equilateral triangle.

An intriguing open question motivated by our work is the following (Question~\ref{ques:weak-strong}). Given a Jordan domain $D$ and two continuous injective curves $\alpha,\beta$ with image in the boundary $\partial D$, is the weak geodesic Fr\'{e}chet distance between $\alpha$ and $\beta$ equal to the strong geodesic Fr\'{e}chet distance? The weak version of the Fr\'{e}chet distance allows $\alpha$ and $\beta$ to be reprarametrized non-injectively, whereas the strong version does not.

We review related work in Section~\ref{sec:related}, state our problem of interest in Section~\ref{sec:preliminaries}, and describe some basic properties of the sweeping cost in Section~\ref{sec:properties}. In Section~\ref{sec:lower-bound} we provide a lower bound on the sweeping cost in terms of shortest area-bisecting curves. We provide analytic formulas for the sweeping costs of Jordan and convex domains in Sections~\ref{sec:jordan} and \ref{sec:convex}, and in Section~\ref{sec:extremal} we deduce that the sweeping cost of a unit-area convex domain is maximized by the equilateral triangle. The conclusion describes related problems of interest, and the appendix contains two technical lemmas and their proofs.

\section{Related work}\label{sec:related}

A wide variety of pursuit-evasion problems have appeared in the mathematics, computer science, engineering, and robotics literature; see \cite{chung2011search} for a survey. Space can modeled in a discrete fashion, for example by a graph \cite{alspach2006searching,bienstock1991graph}, or as a continuous domain in Euclidean space as we consider here. Time can similarly be discrete (turn-based) or continuous, as is our case. See \cite{alonso1992lion,beveridge2015two,chin2010detection,cortes2002coverage,liu2005mobility,parsons1978pursuit} for a selection of such problems.

A further important distinction in a pursuit-evasion problem is whether information is complete (pursuers and intruders know each others' locations), incomplete (pursuers and intruders are invisible to each other), or somewhere in-between. Our problem can be considered as one in which the pursuer has no knowledge of the intruders' movements, whereas the intruders have complete knowledge of the pursuer's current position and future movements. In other words, the pursuer must catch every possible intruder. Evasion problems in which the pursuer has no information and the intruders have complete information can equivalently be cast as contamination-clearing problems, see for example \cite{berger2009many,de2006coordinate,adams2015evasion}. Indeed, the contaminated region of the domain at a particular time includes all locations where an intruder could currently be located, and the uncontaminated region is necessarily free of intruders. It is the task of the pursuer to clear the entire domain of contamination, so that no possible intruders could remain undetected.

The paper \cite{efrat2002new} introduces the \emph{geodesic width} between two polylines, a notion that is very relevant for our problem. The geodesic width between two curves $\alpha$, $\beta$ is the same as the strong geodesic Fr\'{e}chet distance between them (Definition~\ref{def:geodesic-frechet}) when domain $D$ is chosen to be a region with boundary consisting of curves $\alpha$, $\beta$, and the two shortest paths connecting the endpoints of $\alpha$ and $\beta$. If curves $\alpha$ and $\beta$ are polylines with $n$ vertices in total, then \cite{efrat2002new} gives an $O(n^2\log n)$ algorithm for computing the geodesic width between them. The goal of our paper is instead to rigorously prove an analytic formula for the sweeping cost of a domain, Theorem~\ref{thm:frechet}, that is closely related to geodesic widths. Indeed, the right hand side of \eqref{eq:frechet} in Theorem~\ref{thm:frechet} is unchanged if we replace the geodesic distance between two curves (Definition~\ref{def:geodesic-dist-curves}) with the weak geodesic Fr\'{e}chet distance between them (Definition~\ref{def:geodesic-frechet}). The paper \cite{efrat2002new} also studies sweeps of planar domains by piecewise linear curves in which the cost of a sweep is not equal to a length, but instead to the number of vertices or joints in the curve.

Related notions to the geodesic width include the isotopic Fr\'{e}chet distance \cite{chambers2011isotopic} and the minimum deformation area \cite{wangmeasuring} between two curves $\alpha$ and $\beta$. Whereas the geodesic width considers deformations between $\alpha$ and $\beta$ such that no two intermediate curves intersect, this restriction is not present for the isotopic Fr\'{e}chet distance, which can therefore be defined between intersecting curves. The paper \cite{wangmeasuring} considers a distance between two curves on a 2-manifold which is instead an area: the minimal total surface area swept out by any deformation between the two curves. If the curves are piecewise linear in the plane, have $n$ total vertices, and have $I$ intersection points, then \cite{wangmeasuring} gives an $O(n+I^2\log n)$ algorithm to compute the minimum deformation area between them.

\section{Preliminaries and notation}\label{sec:preliminaries}

Let $d\colon\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ denote the Euclidean metric on $\mathbb{R}^2$. The distance between two subsets $X,Y\subseteq\mathbb{R}^2$ is defined as $d(X,Y)=\inf\{d(x,y)~|~x\in X\mbox{ and }y\in Y\}$. We denote the closure of a set $X\subseteq\mathbb{R}^2$ by $\overline{X}$.

\subsection*{Jordan domains and geodesics}

Let $D\subseteq\mathbb{R}^2$ be a Jordan domain, i.e.\ the homeomorphic image of a closed disk in $\mathbb{R}^2$. It follows that $D$ is compact and simply-connected, and its boundary $\partial D$ is a topological circle. Given a point $x\in D$, we let $B(x,\epsilon)=\{y\in D~|~d(x,y)<\epsilon\}$ denote the open ball about $x$ in $D$. We denote the $\epsilon$-offset of a set $X\subseteq D$ by $B(X,\epsilon)=\cup_{x\in X}B(x,\epsilon)$. Given a subset $X\subseteq D$, we define its boundary as $\partial X=\overline{X}\cap\overline{\mathbb{R}^2\setminus X}$.

We refer the reader to \cite{burago2001course} for the basics of geodesic curves and distances. The length of a continuous path $\gamma\colon[a,b]\to D$ is defined as in \cite[Defintion~2.3.1]{burago2001course}; we denote this length by $L(\gamma)$. Curve $\gamma$ is said to be \emph{rectifiable} if $L(\gamma)<\infty$. Domain $D$ has a \emph{length structure} (\cite[Section~2.1]{burago2001course}) in which all continuous paths are admissible, and the length is given by the function $L$.
The associated \emph{geodesic metric} $d_L\colon D\times D\to\mathbb{R}$, also known as a \emph{path-length} or \emph{intrinsic metric}, is
\[ d_L(x,y)=\inf\{L(\gamma)~|~\gamma\colon[a,b]\to D\mbox{ is continuous with }\gamma(a)=x,\ \gamma(b)=y\}. \]
The precise definition of a geodesic, or length-minimizing curve in $D$, is given in \cite[Definition~2.5.27]{burago2001course}. Since $D$ is a Jordan domain, it follows from \cite{bourgin1989shortest,bishop2005intrinsic} that each pair of points in $D$ is joined by a unique shortest geodesic in $D$.

\begin{definition}\label{def:geodesic-dist-curves}
Let $D\subseteq \mathbb{R}^2$ be a Jordan domain. We define the geodesic distance between two curves $\alpha,\beta\colon [a,b]\to D$ to be
\[ d_L(\alpha,\beta)=\max_{t\in[a,b]}d_L(\alpha(t),\beta(t)). \]
\end{definition}

\subsection*{Fr\'{e}chet and geodesic Fr\'{e}chet distances}

The \emph{Fr\'{e}chet distance} is a measure of similarity between two curves $\alpha, \beta \colon [0,1] \to \mathbb{R}^2$. One application of the Fr\'{e}chet distance is in handwriting input recognition for a computer \cite{sriraghavendra2007frechet}: in order to properly tell which letters a user has written, the machine must determine which curves (representing letters) are the most similar. Other notions of distance, such as the Hausdorff distance between the images of the curves, are not necessarily sensitive enough for this task.

The intuition behind the Fr\'{e}chet distance is that you are walking along path $\alpha$, your dog is walking along path $\beta$, and you want to know how long of a leash you need. There are two notions, namely the \emph{weak Fr\'{e}chet distance} and the \emph{strong Fr\'{e}chet distance}. In the weak case, you and your dog are allowed to backtrack along your respective paths, but in the strong case backtracking is forbidden. In general these two distances need not be equal.

\begin{definition}
Let $\alpha, \beta \colon [0,1] \to \mathbb{R}^2$ be continuous curves. Then the \emph{weak (resp.\ strong) Fr\'{e}chet distance} between $\alpha$ and $\beta$ is
\[d_{\mbox{Fr\'{e}chet}}(\alpha,\beta) = \inf_{a,b} \max_{t \in [0,1]} \left\{d\left(\alpha(a(t),\beta(b(t)\right)\right\},\]
where the infimum is taken over all continuous $a, b \colon [0,1] \to [0,1]$ which are surjective (resp.\ bijective).
\end{definition}

If $D\subseteq \mathbb{R}^2$ is a Jordan domain and $\alpha, \beta \colon [0,1] \to D$ are two curves, then we can consider a variant of the Fr\'{e}chet distance in which the Euclidean metric $d$ is replaced with the geodesic metric $d_L$.

\begin{definition}\label{def:geodesic-frechet}
Let $D\subseteq \mathbb{R}^2$ be a Jordan domain, and let $\alpha, \beta \colon [0,1] \to D$ be continuous curves. Then the \emph{weak (resp.\ strong) geodesic Fr\'{e}chet distance} between $\alpha$ and $\beta$ is
\[d_{\mbox{geodesic Fr\'{e}chet}}(\alpha,\beta) = \inf_{a,b} \max_{t \in [0,1]} \left\{d_L\left(\alpha(a(t),\beta(b(t)\right)\right\},\]
where the infimum is taken over all continuous $a, b \colon [0,1] \to [0,1]$ which are surjective (resp.\ bijective).
\end{definition}

\subsection*{Sensor curves}

Let $I=[0,1]$ be the unit interval. We define a sensor curve to be a time-varying rectifiable curve in $D$.

\begin{definition}\label{def:sensor-curve}
A \emph{sensor curve} is a continuous map $f\colon I\times I\to D$ such that
\begin{enumerate}
\item[(i)] each curve $f(\cdot,t)\colon I\to D$ is rectifiable and injective for $t\in(0,1)$,
\item[(ii)] $f(I,0)$ and $f(I,1)$ are each (possibly distinct) single points in $\partial D$, and
\item[(iii)] $f(s,t)\in\partial D$ implies $s\in\{0,1\}$ or $t\in\{0,1\}$.
\end{enumerate}
\end{definition}
\noindent
We think of the first input $s$ as a spatial variable and of the second $t$ as a temporal variable; in particular $f(I,t)$ is the region covered by the curve of sensors at time $t$. Assumption (ii) states that the images of the sensor curve at times 0 and 1 are single points, and assumption (iii) implies that (apart from times 0 and 1) only the boundary of the sensor curve intersects $\partial D$. We define the length of a sensor curve to be
\[ L(f)=\max_{t\in I}\ L(f(\cdot,t)). \]

An \emph{intruder} is a continuous path $\gamma\colon I\to D$. We say that an intruder is caught by a sensor curve $f$ at time $t$ if $\gamma(t)\in f(I,t)$. A path $\gamma\colon[0,t]\to D$ such that $\gamma(t')\notin f(I,t')$ for all $t'\in[0,t]$ is called an \emph{evasion path}. Sensor curve $f$ is a \emph{sweep} if every continuously moving intruder $\gamma\colon I\to D$ is necessarily caught at some time $t$, or equivalently, if no evasion path over the full time interval $I$ exists.\footnote{Our definition is similar to the graph-based definition in \cite[Definition~2.1]{alspach2006searching}.}

The following notation will prove convenient. Fix a sensor curve $f$. We let $C(t)\subseteq D$ be the \emph{contaminated region} at time $t$, and we let $U(t)\subseteq D$ be the \emph{uncontaminated region} at time $t$. More precisely,
\[ C(t)=\{x\in D~|~\exists~\gamma\colon[0,t]\to D \mbox{ with }\gamma(t)=x\mbox{ and }\gamma(t')\notin f(I,t')~\forall t'\in[0,t]\}\quad\mbox{and}\quad U(t)=D\setminus C(t). \]
Note that sensor curve $f$ is a sweep if and only if $C(1)=\emptyset$, or equivalently $U(1)=D$.

\begin{definition}
Let $\mathcal{F}(D)$ be the set of all sensor curve sweeps of $D$. The \emph{sweeping cost} of $D$ is
\[ \mathrm{SC}(D)=\inf_{f\in\mathcal{F}(D)}\ L(f).\]
\end{definition}

\begin{remark}
The results of Sections~\ref{sec:properties}--\ref{sec:lower-bound} hold even if assumptions (ii) and (iii) in Definition~\ref{def:sensor-curve} are removed.
\end{remark}

\section{Properties of sensor sweeps}\label{sec:properties}

We now prove some basic properties of sensor sweeps and the contaminated and uncontaminated regions.

\begin{lemma}\label{lem:connected}
If $x$ and $x'$ are in the same path-connected component of $D\setminus f(I,t)$, then $x\in U(t)$ if and only if $x'\in U(t)$.
\end{lemma}

\begin{proof}
Suppose for a contradiction that $x\in U(t)$ but $x'\notin U(t)$. Since $x'\in C(t)$, there exists an evasion path $\gamma : [0,t] \to D$ with $\gamma(t)=x'$. Since $x$ and $x'$ are in the same path-connected component, there exists a path $\beta\colon I \to D\setminus f(I,t)$ with $\beta(0)=x$ and $\beta(1)=x'$.

Note that $\beta(I)$ and $f(I,t)$ are compact, since they are each a continuous image of the compact set $I$. As any metric space is normal, there exist disjoint neighborhoods containing $\beta(I)$ and $f(I,t)$. Because $f$ is uniformly continuous (it is a continuous function on a compact set), we can choose $\delta_1 > 0$ such that $f(I,[t-\delta_1,t])$ remains in this open neighborhood disjoint from $\beta(I)$, giving
\begin{equation}\label{eq:beta}
\beta(I) \cap f(I,[t-\delta_1,t])=\emptyset.
\end{equation}

Since metric space $D$ is normal, there exist disjoint neighborhoods containing $\gamma(t)=x'$ and $f(I,t)$. Since $\gamma$ is continuous and $f$ is uniformly continuous, we can choose $\delta_2 > 0$ such that $\gamma([t-\delta_2,t])$ and $f(I,[t-\delta_2,t])$ remain in these disjoint neighborhoods, giving
\begin{equation}\label{eq:gamma}
\gamma([t-\delta_2,t]) \cap f(I,[t-\delta_2,t])=\emptyset
\end{equation}

Let $\delta=\min\{\delta_1,\delta_2\}$. Using \eqref{eq:beta} and \eqref{eq:gamma} we can define an evasion path $\gamma\colon[0,t]\to D$ with $\gamma(t)=x$. Indeed, let
\[ \gamma(t')=\begin{cases}
\gamma(t')&\mbox{if }t'\le t-\delta\\
\gamma(2t'-t+\delta)&\mbox{if }t-\delta<t'\le t-\frac{\delta}{2}\\
\beta(\frac{2}{\delta}(t'-t+\frac{\delta}{2}))&\mbox{if }t-\frac{\delta}{2}<t'\le t.
\end{cases} \]
This contradicts the fact that $x\in U(t)$.
\end{proof}

\begin{lemma}\label{lem:closed-open}
 For all $t\in I$, the set $U(t)$ is closed and the set $C(t)$ is open in $D$.
\end{lemma}

\begin{proof}
Suppose $x\in C(t)$. Since $f(I,t)$ is closed and $x\notin f(I,t)$, there exists some $\varepsilon > 0$ such that $B(x,\varepsilon) \cap f(I,t) = \emptyset.$ Note all $x'\in B(x,\varepsilon)$ are in the same path-connected component of $D\setminus f(I,t)$ as $x$ via a straight line path. Hence Lemma~\ref{lem:connected} implies $B(x,\varepsilon)\subseteq C(t)$, showing $C(t)$ is open in $D$. It follows that $U(t)=D\setminus C(t)$ is closed in $D$.
\end{proof}

\begin{lemma}\label{lem:homeo}
If $h\colon D\to h(D)$ is a homeomorphism onto its image $h(D)\subseteq \mathbb{R}^2$, then a sensor curve $f\colon I\times I\to D$ is a sweep of $D$ if and only if sensor curve $hf\colon I\times I\to h(D)$ is a sweep of $h(D)$.
\end{lemma}

\begin{figure}[h]
	\def6.5in{3in}
	\import{}{homeomorphism.pdf_tex}
\caption{A homeomorphism $h\colon D\to h(D)$.}
\label{fig:homeomorphism}
\end{figure}

\begin{proof}
Note that if $\gamma\colon I\to D$ is an evasion path for $f$, then $h\gamma\colon I\to h(D)$ is an evasion path for $hf$. Conversely, if $\gamma\colon I\to h(D)$ is an evasion path for $hf$, then $h^{-1}\gamma\colon I\to D$ is an evasion path for $f$.
\end{proof}

\section{A lower bound on the sweeping cost}\label{sec:lower-bound}

In this section we prove that the sweeping cost of a Jordan domain is at least as large as the length of the shortest area-bisecting curve. The first lemma is a version of the intermediate value theorem with slightly relaxed hypotheses.

\begin{lemma}\label{lem:gen-intermediate-value}
If $f\colon[a,b]\to\mathbb{R}$ is upper semi-continuous and left continuous and if $f(a)<u<f(b)$, then there exists some $c\in(a,b)$ with $f(c)=u$.
\end{lemma}

\begin{proof}
Let $S$ be the set of all $x\in(a,b)$ with $f(x)< u$. Then $S$ is nonempty since $a\in S$, and $S$ is bounded above by $b$. Hence by the completeness of $\mathbb{R}$, the supremum $c=\sup S$ exists. We claim that $f(c)=u$.

Let $\epsilon>0$. Since $f$ is left-continuous, there is some $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ whenever $x\in(c-\delta,c]$. By the definition of supremum, there exists some $y\in(c-\delta,c]$ that is contained in $S$, giving $f(c)<f(y)+\epsilon<u+\epsilon$. Since this is true for all $\epsilon>0$, it follows that $f(c)\le u$.

It remains to show $f(c)\ge u$. Let $\epsilon>0$. Since $f$ is upper semi-continuous, there exists a $\delta>0$ such that $f(c)>f(x)-\epsilon$ whenever $x\in(c-\delta,c+\delta)$. Let $y\in(c,c+\delta)$ and note that $y\notin S$, giving $f(c)>f(y)-\epsilon\ge u-\epsilon$. It follows that $f(c)\ge f(u)$.
\end{proof}

If $S\subseteq \mathbb{R}^2$ is a measurable set, then we let $\mathrm{area}(S)$ denote its area.

\begin{lemma}\label{lem:uppersemi-left-cont}
The function $\mathrm{area}(U(t))$ is upper semi-continuous and left continuous.
\end{lemma}

\begin{proof}
Let $t_0\in I$. We will show that $\mathrm{area}(U(t))$ is right upper semi-continuous and left continuous at $t_0$, which implies the function is both upper semi-continuous and left continuous.

For right upper semi-continuity, note for $t\ge t_0$ we have
\begin{equation}\label{eq:contain}
U(t)\subseteq U(t_0)\cup f(I,[t_0,t]).
\end{equation}
Since sensor curve $f\colon I\times I\to D$ is a continuous function on a compact domain, it is also uniformly continuous. Hence for all $\epsilon>0$ there exists some $\delta$ such that
\begin{equation}\label{eq:in-ball}
f(I,[t_0,t_0+\delta])\subseteq B(f(I,t_0),\epsilon).
\end{equation}
It follows that for all $t\in[t_0,t_0+\delta]$ we have
\begin{align*}
\mathrm{area}(U(t))-\mathrm{area}(U(t_0))&\le\mathrm{area}(f(I,[t_0,t]))&&\mbox{by \eqref{eq:contain}}\\
&\le\mathrm{area}(B(f(I,t_0),\epsilon))&&\mbox{by \eqref{eq:in-ball}}\\
&\le 2L(f(I,t_0))\epsilon+\pi\epsilon^2,
\end{align*}
where the last inequality is by a result of Hotelling (see for example \cite[Equation~(2.1)]{johnstone1989hotelling}). Hence $\mathrm{area}(t)$ is right upper semi-continuous.

To see that $\mathrm{area}(U(t))$ is left continuous at $t_0\in I$, we must show that for all sequences $\{s_i\}$ with $0\le s_i\le t_0$ and $\lim_i s_i=t_0$, we have $\lim_i\mathrm{area}(U(s_i))=\mathrm{area}(U(t_0))$.
We claim
\begin{equation}\label{eq:containments}
U(t_0)\setminus f(I,t_0)\subseteq\liminf_i U(s_i)\subseteq\limsup_i U(s_i)\subseteq U(t_0),
\end{equation}
where the middle containment is by definition.
We now justify the first and last containment.

To prove $U(t_0)\setminus f(I,t_0)\subseteq\liminf_i U(s_i)$ it suffices to show that for any $x \in U(t_0) \setminus f(I,t_0)$ there exists an $\epsilon >0$ such that $x \in U(t_0 - \delta)$ for all $\delta \in [0,\epsilon)$.  Fix $\epsilon$ such that $x \notin f(I,t)$ for $t \in (t_0-\epsilon, t_0]$. Suppose for a contradiction that $x\in C(t_0-\delta)$ for some $\delta \in [0,\epsilon)$.  Hence there exists an evasion path $\gamma \colon [0,t_0-\delta] \to D$ with $\gamma(t_0-\delta) = x$. It is possible to extend $\gamma$ to an evasion path $\tilde{\gamma}\colon[0,t_0]\to D$ defined by
\[ \tilde{\gamma}(t)=\begin{cases}
\gamma(t)&\mbox{if }t\in[0,t_0-\delta]\\
x&\mbox{if }t\in(t_0-\delta,t_0].
\end{cases} \]
This contradicts the fact $x \in U(t_0)$, thus giving the first containment.

We now show $\limsup_i U(s_i)\subseteq U(t_0)$. If $x\notin U(t_0)$, then there exists an evasion path $\gamma\colon[0,t_0]\to D$ with $\gamma(t_0)=x$. Since $C(t_0)$ is open by Lemma~\ref{lem:closed-open}, there exists some $\delta>0$ such that $B(x,\delta)\subseteq C(t_0)$, and hence $B(x,\delta)\cap f(I,t_0)=\emptyset$. Since $f$ is uniformly continuous, there is some $\epsilon_1>0$ sufficiently small with $B(x,\delta)\cap f(I,[t_0-\epsilon_1,t_0])=\emptyset$, and since $\gamma$ is continuous there is some $\epsilon_2>0$ with $\gamma([t_0-\epsilon_2,t_0])\subseteq B(x,\delta)$. Let $\epsilon=\min\{\epsilon_1,\epsilon_2\}$. Reparametrize $\gamma$ to get a continuous curve $\tilde{\gamma}\colon[0,t_0]\to D$ with
\[ \tilde{\gamma}(t)=\begin{cases}
\gamma(t)&\mbox{if }x\in[0,t_0-\epsilon)\\
\gamma(t)\in B(x,\delta)&\mbox{if }t\in[t_0-\epsilon,t_0-\epsilon/2)\\
\gamma(t)=x&\mbox{if }t\in[t_0-\epsilon/2,t_0].
\end{cases} \]
The evasion path $\tilde{\gamma}$ shows $x\notin\limsup_i U(s_i)$, giving the third containment and finishing the proof of \eqref{eq:containments}.

Set $f(I,t_0)$ has Lebesgue measure zero since curve $f(\cdot,t_0)$ is rectifiable, giving $\mathrm{area}(U(t_0)\setminus f(I,t_0))=\mathrm{area}(U(t_0))$. Thus \eqref{eq:containments} implies
\[ \mathrm{area}(\limsup_i U(s_i))=\mathrm{area}(U(t_0))=\mathrm{area}(\liminf_i U(s_i)). \]
Since $\mathrm{area}(D)$ is finite, Lemma~\ref{lem:measures} implies $\limsup_i\mathrm{area}(U(s_i))\le\mathrm{area}(\limsup_i U(s_i))$ and $\mathrm{area}(\liminf_i U(s_i))\le\liminf_i\mathrm{area}(U(s_i))$, giving
\[ \limsup_i\mathrm{area}(U(s_i))\le\mathrm{area}(U(t_0))\le\liminf_i\mathrm{area}(U(s_i)). \]
Hence $\lim_i\mathrm{area}(U(s_i))=\mathrm{area}(U(t_0))$ as required.
\end{proof}

\begin{remark}
The function $\mathrm{area} (U(t))$ need not be right continuous. Indeed, consider a sensor curve as shown below in Figure~\ref{fig:failed_sweep}, where $\mathrm{area}(U(t_0))>0$, and where there is some $\epsilon_0>0$ such that for all $0<\epsilon<\epsilon_0$, only one point on the sensor curve at time $t_0+\epsilon$ intersects $\partial D$ and $\mathrm{area}(U(t_0+\epsilon))=0$.

\begin{figure}[h]
	\def6.5in{6.5in}
	\import{}{failed_sweep.pdf_tex}
\caption{An example sensor curve where the function $\mathrm{area}(U(t))$ is not right continuous. The shaded region is $U(t)$ and the unshaded region is $C(t)$.}
\label{fig:failed_sweep}
\end{figure}
\end{remark}

As a consequence we obtain the following lower bound on the sweeping cost.

\begin{theorem}\label{thm:area-lower-bound}
If $D$ is a Jordan domain, then the sweeping cost $\mathrm{SC}(D)$ is at least as large as the length of the shortest area-bisecting curve in $D$.
\end{theorem}

\begin{proof}
Suppose that $f$ is a sweep of $D$. Note that $\mathrm{area}(U(0))=0$ and $\mathrm{area}(U(1))=\mathrm{area}(D)$. By Lemmas~\ref{lem:gen-intermediate-value} and \ref{lem:uppersemi-left-cont}, there exists some time $t'\in I$ with $\displaystyle \mathrm{area}(U(t'))=\tfrac{1}{2}\mathrm{area}(D)$. So $L(f(\cdot,t'))$ and hence $\mathrm{SC}(D)$ is at least as large as the shortest area-bisecting curve in $D$.
\end{proof}

\begin{example}\label{example:disk}
If $D = \{(x,y)\in\mathbb{R}^2~|~x^2+y^2\le1\}$ is the unit disk, then $\mathrm{SC}(D)=2$.
\end{example}

\begin{proof}
To see $\mathrm{SC}(D)\le2$, consider the sweep $f\colon[-1,1]\times I\to D$ defined by $f(s,t)=(2s \sqrt{t-t^2},2t-1)$ (Figure~\ref{fig:disk-sweep}) which has length $2$.

\begin{figure}[h]
	\def6.5in{6.5in}
	\import{}{circle.pdf_tex}
\caption{A sweep of the unit disk. The shaded region is $U(t)$ and the unshaded region is $C(t)$.}
\label{fig:disk-sweep}
\end{figure}

For the reverse direction, note that the shortest area-bisecting curve in $D$ is a diameter \cite{esposito2012longest}. Hence we apply Theorem~\ref{thm:area-lower-bound} to get $\mathrm{SC}(D)\ge2$.
\end{proof}

\begin{example}
Let $a,b>0$. If $D=\{(x,y)\in\mathbb{R}^2~|~(x/a)^2+(y/b)^2\le1\}$ is the convex hull of an ellipse, then $\mathrm{SC}(D)=\min\{2a,2b\}$.
\end{example}

\begin{proof}
To see $\mathrm{SC}(D)\le\min\{2a,2b\}$, construct a sweep much like in Example~\ref{example:disk}.

For the reverse direction, the solution to \cite[Chapter~X, Problem~33]{polya1965mathematics} states that because the ellipse has a center of symmetry, the shortest area-bisecting curve is a straight line. All area-bisecting lines pass through the center of the ellipse, and hence have length at least $\min\{2a,2b\}$.
It follows from Theorem~\ref{thm:area-lower-bound} that $\mathrm{SC}(D)\ge\min\{2a,2b\}$.
\end{proof}

\section{A lemma of no progress}

In Sections~\ref{sec:jordan}--\ref{sec:extremal} we will restrict attention to sensor curves $f$ with boundary points $f(0,t),f(1,t)\in\partial D$ for all $t\in I$. The motivation behind this assumption is Lemma~\ref{lem:no-progress}, which states that if $f(0,t)\notin\partial D$ or $f(1,t)\notin\partial D$, then the uncontaminated region at time $t$ is as small as possible, namely $U(t)=f(I,t)$.

The following lemma is from \cite{zoretti1905fonctions}; see also its statement in \cite[page 164]{kline1928separation}.

\begin{lemma}[Zoretti]\label{lem:zoretti}
If $K$ is a bounded maximal connected subset of a plane closed set $M$ and $\epsilon>0$, then there exists a simple closed curve $J$ enclosing $K$ such that $J\cap M=\emptyset$ and $J\subseteq B(K,\epsilon)$.
\end{lemma}

\begin{lemma}\label{lem:no-progress}
Let $f\colon I\times I\to D$ be a sensor curve. If $f(0,t)\notin\partial D$ or $f(1,t)\notin\partial D$ and $U(t)\neq D$, then $U(t)=f(I,t)$.
\end{lemma}

\begin{proof}
Without loss of generality suppose $f(0,t)\notin \partial D$. It suffices to show that $D\setminus f(I,t)$ is a single path-connected component, because then Lemma~\ref{lem:connected} and the fact that $U(t)\neq D$ will imply $C(t)=D\setminus f(I,t)$ and hence $U(t)=f(I,t)$. Let $x,x'\in D\setminus f(I,t)$; we must find a path in $D\setminus f(I,t)$ connecting $x$ and $x'$. There are two cases: when $f(1,t)\notin\partial D$, and when $f(1,t)\in\partial D$.

In the first case $f(1,t)\notin\partial D$, note that $f(I,t)$ is disjoint from $\partial D$. Hence by compactness there exists some $\epsilon>0$ such that $d(f(I,t),\partial D\cup\{x,x'\})<\epsilon$. By Lemma~\ref{lem:zoretti} (with $K=f(I, t)$ and $M=\partial D\cup\{x,x'\}$), there exists a simple closed curve $J$ in $D$ enclosing $f(I, t)$ but not enclosing $x$ or $x'$. We may therefore connect $x$ and $x'$ by a path in $D\setminus f(I,t)$ consisting of three pieces: a path in $D$ from $x$ to $J$, a path in $D$ from $x'$ to $J$, and a path in $J$ connecting these two endpoints (Figure~\ref{fig:no-progress-1}).

\begin{figure}[h]
	\def6.5in{6.5in}
	\import{}{no_progress_1.pdf_tex}
\caption{The first case in the proof of Lemma~\ref{lem:no-progress}, with $J$ drawn in red.}
\label{fig:no-progress-1}
\end{figure}

In the second case $f(1,t)\in\partial D$, pick some point $y\in D\setminus(f(I,t)\cup\{x,x'\})$. By translating $D$ in the plane we may assume that $y=\vec{0}$. Define the inversion function $i\colon\mathbb{R}^2\setminus\{\vec{0}\}\to\mathbb{R}^2\setminus\{\vec{0}\}$ by $i(r\cos\theta,r\sin\theta)=(\frac{1}{r}\cos\theta,\frac{1}{r}\sin\theta)$; note $i^2$ is the identity map. Let $\epsilon>0$ be such that
\[d(i(f(I, t)\cup \partial D),\{i(x),i(x')\})<\epsilon.\] By Lemma~\ref{lem:zoretti} (with $K=i(f(I, t)\cup \partial D)$ and $M=\{i(x),i(x')\}$), there exists a simple closed curve $J$ in $\mathbb{R}^2$ enclosing $i(f(I, t)\cup \partial D)$ but not enclosing $i(x)$ or $i(x')$. By the Jordan curve theorem, $i(x)$ and $i(x')$ are in the same (exterior) connected component $E$ of $\mathbb{R}^2\setminus J$. Since $E$ is open it is also path-connected,
and hence we can connect $i(x)$ and $i(x')$ by a path $\gamma$ in $E$. The path $i(\gamma)$ is therefore a path in $D\setminus f(I,t)$ connecting $x$ and $x'$ (Figure~\ref{fig:no-progress-2}).

\begin{figure}[h]
	\def6.5in{6.5in}
	\import{}{no_progress_2.pdf_tex}
\caption{The second case in the proof of Lemma~\ref{lem:no-progress}, with $J$ and $i(J)$ drawn in red.}
\label{fig:no-progress-2}
\end{figure}
\end{proof}

\section{Sweeping cost of a Jordan domain}\label{sec:jordan}

As motivated by Lemma~\ref{lem:no-progress}, for the remainder of the paper we restrict attention to sensor curves satisfying $f(s,t)\in\partial D$ if and only if $s\in\{0,1\}$ or $t\in\{0,1\}$.

Given curves $\alpha,\beta\colon I\to\partial D$ with $\alpha(1)=\beta(0)$ (see Figure~\ref{fig:alpha_beta_winding}), we define the concatenated curve $\alpha\cdot\beta\colon I\to\partial D$ by
\[ \alpha\cdot\beta(t)=\begin{cases}
\alpha(2t) &\mbox{if }0\le t\le \frac{1}{2}\\
\beta(2t-1) &\mbox{if }\frac{1}{2}<t\le 1.
\end{cases} \]
We define the inverse curve $\beta^{-1}\colon I\to\partial D$ by $\beta^{-1}(t)=\beta(1-t)$.

\begin{figure}[h]
	\def6.5in{1.5in}
	\import{}{alpha_beta.pdf_tex}
	\hspace{15mm}
	\def6.5in{4in}
	\import{}{winding_number.pdf_tex}
\caption{(Left) Two curves $\alpha,\beta\colon I\to\partial D$ with $\alpha(1)=\beta(0)$. (Right) Example winding numbers.}
\label{fig:alpha_beta_winding}
\end{figure}

Since $\partial D$ is homeomorphic to the circle, given a loop $\gamma\colon I\to\partial D$ (with $\gamma(0)=\gamma(1)$) we can denote the winding number of $\gamma$, i.e.\ the number of times $\gamma$ wraps around $\partial D$, by $\mathrm{wn}(\gamma)$. The winding number is positive (resp.\ negative) for loops that wrap around in the counterclockwise (resp.\ clockwise) direction. Note that if $\alpha$ and $\beta$ are paths in $\partial D$ with $\alpha(0)=\beta(0)$ and $\alpha(1)=\beta(1)$, then $\alpha\cdot\beta^{-1}$ is a loop.

Our main result is an analytic formula for the sweeping cost of a Jordan domain.

\begin{theorem}\label{thm:frechet}
The sweeping cost of a Jordan domain $D$ is
\begin{equation}\label{eq:frechet}
\mathrm{SC}(D)=\inf\{d_L(\alpha,\beta)~|~\alpha,\beta\colon I\to\partial D,\ \alpha(0)=\beta(0),\ \alpha(1)=\beta(1),\ \mathrm{wn}(\alpha\cdot\beta^{-1})\neq0\}.
\end{equation}
\end{theorem}

Equation~\eqref{eq:frechet} is closely related to the \emph{geodesic width} between two polylines \cite{efrat2002new}, and also the \emph{isotopic Fr\'{e}chet distance} between two curves \cite{chambers2011isotopic}. Indeed, note that the right hand side of \eqref{eq:frechet} is unchanged if we replace $d_L(\alpha,\beta)$ with the weak geodesic Fr\'{e}chet distance
between $\alpha$ and $\beta$ (see Definition~\ref{def:geodesic-frechet}).

\begin{remark}\label{rem:frechet-unchanged}
The value of the right hand side of \eqref{eq:frechet} is unchanged if we replace $\mathrm{wn}(\alpha\cdot\beta^{-1})\neq0$ with $\mathrm{wn}(\alpha\cdot\beta^{-1})\in\{-1,1\}$.
\end{remark}

\begin{proof}[Proof of Remark~\ref{rem:frechet-unchanged}]
Let $\alpha,\beta\colon I\to\partial D$ with $\alpha(0)=\beta(0)$ and $\alpha(1)=\beta(1)$, and suppose $|\mathrm{wn}(\alpha\cdot\beta^{-1})|\ge2$.
Hence there exists some $0<t<1$ such that $\alpha(t)=\beta(t)$ and $\mathrm{wn}(\alpha|_{[0,t]}\cdot\beta|_{[0,t]}^{-1})\in\{-1,1\}$.
The claim follows since
\[ d_L(\alpha|_{[0,t]},\beta|_{[0,t]})\le d_L(\alpha,\beta). \]
\end{proof}

The following lemma will be used to prove the $\le$ direction in \eqref{eq:frechet}.

\begin{lemma}\label{lem:frechet-le}
Let $D$ be a Jordan domain. Suppose $\alpha,\beta\colon I\to\partial D$ with $\alpha(0)=\beta(0)$, $\alpha(1)=\beta(1)$, and $\mathrm{wn}(\alpha\cdot\beta^{-1})\neq 0$. If $f \colon I \times I \to D$ is any sensor curve with $f(0,t)=\alpha(t)$ and $f(1,t)=\beta(t)$, then $f$ is a sweep of $D$.
\end{lemma}

\begin{proof}
Let $\mathbb{D}=\{(x,y)\in\mathbb{R}^2~|~x^2+y^2\le1\}$ be the unit disk. We first prove this claim in the case when $D=\mathbb{D}$.

By Lemma~\ref{lem:disk-curves}, there exists a point $p\in\mathbb{R}^2\setminus\mathbb{D}$ and two continuous families of curves $g_\alpha,g_\beta\colon I\times I\to\mathbb{R}^2$ such that
\begin{itemize}
\item $g_\alpha(0,t)=p=g_\beta(0,t)$,
\item $g_\alpha(1,t)=\alpha(t)$,
\item $g_\beta(1,t)=\beta(t)$, and
\item $g_\alpha(s,t),g_\beta(s,t)\notin\mathbb{D}$ for $s<1$.
\end{itemize}
Let $S^1$ be the circle of unit circumference, i.e.\ $[0,1]$ with endpoints $0$ and $1$ identified. Define a continuous map $g\colon S^1\times I\to D$ via
\[g(s,t)=\begin{cases}
g_\alpha(3st,t)&\mbox{if }0\le s<\frac{1}{3}\\
f(3s-1,t)&\mbox{if }\frac{1}{3}\le s<\frac{2}{3}\\
g_\beta(3t(1-s),t)&\mbox{if }\frac{2}{3}\le s\le 1.\\
\end{cases}\]
Note $g(\cdot,t)$ is indeed a (possibly non-simple) map from the circle since $g_\alpha(0,t)=p=g_\beta(0,t)$ for all $t$. Define a continuous signed distance $d^\pm \colon \mathbb{R}^2 \times I \to \mathbb{R}$ by
\begin{equation*}
d^\pm(x,t)=\begin{cases}
d(x,g(S^1,t)) &\mbox{if }x\in g(S^1,t) \mbox{ or }\mathrm{wn}(g(\cdot,t),x)=0 \\
-d(x,g(S^1,t)) &\mbox{if }x\notin g(S^1,t) \mbox{ and }\mathrm{wn}(g(\cdot,t),x)\neq0.
\end{cases}
\end{equation*}
Here $\mathrm{wn}(g(\cdot,t),x)$ denotes (for $x\notin g(S^1,t)$) the winding number of the map $g(\cdot,t)\colon S^1\to\mathbb{R}^2\setminus\{x\}\simeq S^1$. Note that $\mathrm{wn}(g(\cdot,t),x)$ is constant on each connected component of $\mathbb{R}^2\setminus g(S^1,t)$, and that $d^\pm$ is continuous.

Given any intruder path $\gamma\colon I\to\mathbb{D}$, the continuous function $d^\pm(\gamma(t),t) \colon I \to \mathbb{R}$ satisfies $d^\pm(\gamma(0),0) \ge 0$ (since $f(I,0)$ is a single point in $\partial D$) and $d^\pm(\gamma(1),1) \le 0$ (since $f(I,1)$ is a single point in $\partial D$ and $\mathrm{wn}(\alpha\cdot\beta^{-1})\neq 0$). By the intermediate value theorem there exists some $t' \in I$ with $d^\pm(\gamma(t'),t')=0$, and hence $\gamma(t')\in g(S^1,t')\cap\mathbb{D}=f(I,t')$. So $\gamma$ is not an evasion path, and $f$ is a sweep of $\mathbb{D}$.

We now handle the case when $D$ is an arbitrary Jordan domain. By definition there exists a homeomorphism $h\colon\mathbb{D}\to D$. Note that $h^{-1}\alpha,h^{-1}\beta\colon I\to\partial \mathbb{D}$ with $h^{-1}\alpha(0)=h^{-1}\beta(0)$, $h^{-1}\alpha(1)=h^{-1}\beta(1)$, and $\mathrm{wn}(h^{-1}\alpha\cdot h^{-1}\beta^{-1})\neq 0$. Since $h^{-1}f \colon I \times I \to \mathbb{D}$ is a sensor curve, it follows from our proof in the case of the disk that $h^{-1}f$ is a sweep of $\mathbb{D}$. Hence $f$ is a sweep of $D$ by Lemma~\ref{lem:homeo}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:frechet}]
Let
\[ c = \inf\{d_L(\alpha,\beta)~|~\alpha,\beta\colon I\to\partial D,\ \alpha(0)=\beta(0),\ \alpha(1)=\beta(1),\ \mathrm{wn}(\alpha\cdot\beta^{-1})\neq0\}. \]
We first prove the $\le$ direction of \eqref{eq:frechet}. Let $\epsilon>0$ be arbitrary. By the definition of infimum there exist curves $\alpha,\beta\colon I\to\partial D$ with $\alpha(0)=\beta(0)$, $\alpha(1)=\beta(1)$, $\mathrm{wn}(\alpha\cdot\beta^{-1})\neq0$, and $d_L(\alpha,\beta)\le c+\epsilon$. Define $f\colon I\times I\to D$ by letting $f(\cdot, t)\colon I\to D$ be the unique constant-speed geodesic in $D$ between $f(0,t)=\alpha(t)$ and $f(1,t)=\beta(t)$, which exists by \cite{bourgin1989shortest,bishop2005intrinsic}.
Lemma~\ref{lem:frechet-le} implies that $f$ is a sweep, and hence we have
\[ \mathrm{SC}(D)\le L(f)=d_L(\alpha,\beta)\le c+\epsilon. \]
Since this is true for all $\epsilon>0$, we have $\mathrm{SC}(D)\le c$.

For the $\ge$ direction of \eqref{eq:frechet}, suppose that $f$ is a sensor curve with $L(f)<c$. For notational convenience, define $\alpha,\beta\colon I\to\partial D$ by $\alpha(t)=f(0,t)$ and $\beta(t)=f(1,t)$. Then necessarily $\mathrm{wn}(\alpha\cdot\beta^{-1})=0$, and furthermore
\begin{equation}\label{eq:partial-wn}
\alpha(t)=\beta(t)\quad\mbox{implies}\quad\mathrm{wn}(\alpha|_{[0,t]}\cdot\beta|_{[0,t]}^{-1})=0,
\end{equation}
since otherwise we'd have
\[ L(f)\ge d_L(\alpha,\beta) \ge d_L(\alpha|_{[0,t]},\beta|_{[0,t]})\ge c, \]
a contradiction. We will show that $f$ is not a sweep of $D$ by showing the existence of an evasion path $\gamma\colon I\to D$ whose image furthermore lives in $\partial D$.

Indeed, consider the 1-dimensional evasion problem in $\partial D$ where the region covered by the sensors at time $t$ is $\{\alpha(t),\beta(t)\}$. In this 1-dimensional problem, it is clear that the uncontaminated region in $\partial D$ is either (i) a single point $\alpha(t)=\beta(t)$, (ii) a closed interval in $\partial D$ with endpoints $\alpha(t)$ and $\beta(t)$, or (iii) all of $\partial D$. Equation~\eqref{eq:partial-wn}, however, rules out the possibility of (iii). It follows that the contaminated region in $\partial D$ is always a nonempty open interval in $\partial D$ with continuously varying endpoints $\alpha(t)$ and $\beta(t)$. Therefore we can define an evasion path $\gamma\colon I\to \partial D$, for example by letting $\gamma(t)$ be the midpoint of the open interval of the uncontaminated region in $\partial D$. This evasion path $\gamma$ is also an evasion path for our original 2-dimensional problem in $D$, as $\gamma\colon I\to \partial D\subseteq D$ satisfies $\gamma(t)\notin f(I,t)$ for all $t$. This gives the $\ge$ direction of \eqref{eq:frechet}.
\end{proof}

\begin{question}
Does Theorem~\ref{thm:frechet} hold even if assumption (iii) in Definition~\ref{def:sensor-curve} is removed, i.e.\ if the interior of a sensor curve is also allowed to touch $\partial D$?
\end{question}

\begin{question}\label{ques:weak-strong}
For any Jordan domain $D$ in the plane and injective curves $\alpha,\beta\colon I\to\partial D$,
we conjecture that the weak geodesic Fr\'{e}chet distance between $\alpha$ and $\beta$ is equal to their strong geodesic Fr\'{e}chet geodesic distance.
\end{question}

There are simple counterexamples to Question~\ref{ques:weak-strong} when $\alpha$ and $\beta$ are not injective, or when they do not map to $\partial D$. Closely related is following question: is the value of \eqref{eq:frechet} unchanged if we require $\alpha$ and $\beta$ to be injective?

\section{Sweeping cost of a convex domain}\label{sec:convex}

Given a convex Jordan domain $D\subseteq \mathbb{R}^2$, its \emph{width} $w(D)$ is defined as
\[ w(D) = \min_{\|v\|=1}\ \max_{x\in\mathbb{R}^2}\ L(D\cap\{x+tv~|~t\in\mathbb{R}\}), \]
where $v$ is a unit direction vector in $\mathbb{R}^2$. Alternatively, the width $w(D)$ is the smallest distance between two parallel supporting lines on opposite sides of $D$ (Figure~\ref{fig:width}).

\begin{figure}[h]
	\def6.5in{2in}
	\import{}{width.pdf_tex}
\caption{The width $w(D)$ of a domain $D$.}
\label{fig:width}
\end{figure}

\begin{theorem}\label{thm:sc=width}
If $D$ is a convex Jordan domain, then $\mathrm{SC}(D)=w(D)$.
\end{theorem}

\begin{proof}
We first show $\mathrm{SC}(D)\le w(D)$. By Theorem~\ref{thm:frechet}, it suffices to show
\[ \inf\{d_L(\alpha,\beta)~|~\alpha,\beta\colon I\to\partial D,\ \alpha(0)=\beta(0),\ \alpha(1)=\beta(1),\ \mathrm{wn}(\alpha\cdot\beta^{-1})\neq0\} \le w(D). \]
Let $v$ some direction vector realizing the width, i.e.\ $w(D)=\max_{x\in\mathbb{R}^2}\ L(D\cap\{x+tv~|~t\in\mathbb{R}\})$. Consider sweeping through all lines in $\mathbb{R}^2$ parallel to $v$; the intersection of these lines with $\partial D$ traces out two continuous curves $\alpha,\beta\colon I\to\partial D$ with $\alpha(0)=\beta(0)$, $\alpha(1)=\beta(1)$, and $\mathrm{wn}(\alpha\cdot\beta^{-1})=\pm 1$.\footnote{This is not quite precise if $\partial D$ contains a straight line segment of non-zero length parallel to $v$ (there are at most two such segments). In this case, pick an arbitrary point on each such line segment; each such point will be either the starting point or the ending point for both $\alpha$ and $\beta$.} We have $d_L(\alpha,\beta)\le w(D)$, giving $\mathrm{SC}(D)\le w(D)$.

To finish the proof, we need some background on planar convex domains. A point $x\in\partial D$ is \emph{smooth} if it has a unique supporting hyperplane, and otherwise $x$ is a \emph{vertex} containing a range of angles $[\theta_1,\theta_2]\subseteq S^1$ (with $\theta_1\neq\theta_2$) which are the outward normal directions of supporting hyperplanes of $D$ at $x$. Away from the vertices, the unique supporting hyperplane of $x\in\partial D$ varies continuously with $x$. By \cite[Proposition~11.6.2]{berger2009geometry},
the set of vertices of the closed convex domain $D$ is countable.

We now show $\mathrm{SC}(D)\ge w(D)$. Given $\epsilon>0$, let $\alpha$ and $\beta$ be $\epsilon$-close to realizing the infimum in \eqref{eq:frechet}, meaning $\mathrm{SC}(D)+\epsilon\ge d_L(\alpha,\beta)$. For notational convenience we assume that $\alpha(0)=\beta(0)$ and $\alpha(1)=\beta(1)$ are not vertices of $\partial D$ (our same proof technique works regardless). Let $T=\{t_1,t_2,t_3,\ldots\}\subseteq I$ be a countable subset such that $t\in T$ if either $\alpha(t)$ or $\beta(t)$ is a vertex of $\partial D$. Let $t_0=0$, and if $|T|$ is finite, then let $t_{|T|+1}=1$. For $i=1,2,\ldots,|T|$, choose weights $w_i>0$ such that $\sum_i w_i=w<\infty$; this is possible since $T$ is countable. Let $p_1\colon \partial D \times S^1\to \partial D$ and $p_2\colon \partial D\times S^1\to S^1$ be the projection maps. It is possible to define a \emph{continuous} map $g_\alpha\colon [0,1+w]\to \partial D\times S^1$ satisfying the following properties.
\begin{itemize}
\item Each $g_\alpha(s)$ is equal to a point $(\alpha(t),v)\in\partial D\times S^1$ with $t\in I$ such that $v$ is the outward normal vector to a supporting hyperplane of $D$ at $\alpha(t)$.
\item If $t_i\le t\le t_{i+1}$, then $\alpha(t)=p_1g_\alpha(t+\sum_{j=1}^iw_j)$.
\item For all $0\le s\le w_i$, we have $p_1g_\alpha(s+t_i+\sum_{j=1}^{i-1}w_j)=\alpha(t_i)$.
\item As $s$ varies from $0$ to $w_i$, angle $p_2g_\alpha(s+t_i+\sum_{j=1}^{i-1}w_j)$ varies over the range of supporting hyperplanes of $D$ at $\alpha(t_i)$ (which may be a single angle if $\alpha(t_i)$ is not a vertex of $\partial D$).
\end{itemize}
Define $g_\beta\colon [0,1+w]\to \partial D\times S^1$ similarly (with $\alpha$ replaced everywhere by $\beta$). Note that $g_\alpha(0)=g_\beta(0)$, that $g_\alpha(1+w)=g_\beta(1+w)$, and that $p_2g_\alpha$ and $p_2g_\beta$ wrap in opposite directions around $S^1$. Hence for some $s\in[0,1+w]$ the supporting hyperplanes corresponding to $g_\alpha(s)$ and $g_\beta(s)$ will be parallel and on opposite sides of $D$. It follows that
\[ \mathrm{SC}(D)+\epsilon\ge d_L(\alpha,\beta)\ge d(p_1g_\alpha(s),p_1g_\beta(s))\ge w(D). \]
Since this is true for all $\epsilon>0$, we have $\mathrm{SC}(D)\ge w(D)$.
\end{proof}

The paper \cite{houle1988computing} shows that for $D\subseteq \mathbb{R}^2$ a convex polygonal domain with $n$ vertices, the width and hence the sweeping cost of $D$ can be computed in time $O(n)$ and space $O(n)$ using the \emph{rotating calipers} technique.

\section{Extremal shapes}\label{sec:extremal}

Which convex shape of unit area has the largest sweeping cost? The papers \cite[Theorem~4.3]{cerdan2006comparing} and \cite{pal1921minimumproblem} state that if $D$ is a bounded planar convex domain, then $\mathrm{area}(D)\ge w(D)^2/\sqrt{3}$,
where equality is achieved if $D$ is an equilateral triangle. The next corollary follows immediately from Theorem~\ref{thm:sc=width}.

\begin{corollary}
Let $D$ be a convex Jordan domain. Then
\[ \mathrm{area}(D)\ge\frac{\mathrm{SC}(D)^2}{\sqrt{3}}, \]
where equality is achieved if $D$ is an equilateral triangle. Hence the equilateral triangle has the maximal sweeping cost over all planar convex domains of the same area.
\end{corollary}

The next example shows that there is no extremal shape for non-convex Jordan domains.

\begin{example}
A (non-convex) domain $D$ of unit area may have arbitrarily large sweeping cost.
\end{example}

\begin{proof}
Consider a deformation of an equilateral triangle with unit side lengths where we deform each edge towards the center of the triangle (Figure~\ref{fig:triangle-deform}). Note that as the sweeping cost converges to $\tfrac{1}{\sqrt{3}}$ (the distance from the center to a vertex) from above, the area of the shape tends zero. Rescaling each shape in this deformation to have area one shows that a non-convex domain of unit area may have arbitrarily large sweeping cost.

\begin{figure}[h]
	\def6.5in{6.5in}
	\import{}{triangle_deform.pdf_tex}
\caption{We deform each edge of the triangle towards the center of the triangle, producing a three-pronged shape. As the sweeping costs of the shapes converge to a fixed constant, the areas converge to zero.}
\label{fig:triangle-deform}
\end{figure}
\end{proof}

\section{Conclusion}\label{sec:conclusion}

Given a Jordan domain $D$ in the plane, we show that the sweeping cost of $D$ is at least as large as the shortest area-bisecting curve in $D$, and we give a formula for the sweeping cost in terms of the geodesic Fr\'{e}chet distance between two curves on the boundary of $D$ with non-equal winding numbers. We show that the sweeping cost of any convex domain is equal to its width. Therefore, the sweeping cost of a polygonal convex domain with $n$ vertices can be computed in time and space $O(n)$, and a convex domain of unit area with maximal sweeping cost is the equilateral triangle.

We end by mentioning two related settings of interest. First, let $D$ be a compact region in the plane, perhaps not simply-connected. Suppose the pursuer is now a union of curves. What can one say about the sweeping cost of $D$, measured as the sum of the curve lengths? Second, let $D\subset \mathbb{R}^n$ be the homemorphic image of the closed $n$-dimensional ball. What are the properties of the sweeping cost of $D$, when swept by an $(n-1)$-dimensional ``sensor surface"? For example, what is a 3-dimensional convex body of unit volume which maximizes this higher-dimensional sweeping cost?

\section{Acknowledgements}

We would like to thank Clayton Shonkwiler for pointing us to the references \cite{cerdan2006comparing,pal1921minimumproblem}.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


It's Too Faced's Birthday and they're selling the Too Faced x Skinny Dip London Makeup Bag for $20 USD until supplies run out. They're also offering other products with the makeup bag (see below) for $65, $75 and $100. Personally, I don't think those are worth it! But I love the bag and it's definitely worth $20. Nuthin' flimsy about it. Just throwin' in my two cents...It makes me SMILE :) Had to tell ya!
That make-up bag is so cute, Kim! Admittedly, it's not really my style - my make-up bags are all plain black, ha ha - but I still think it's adorable, especially because it features illustrations of my some of my favourite TooFaced products. (Funnily enough, I was just using my Chocolate Bar palette this morning.) I'm going to have to look up the one with peach print as well, just to see it!
I love the make-up from too faced! Nice make-up bag dear!
I have more makeup bags that I can use, so I am not really tempted, but I tried some Too Faced blushes recently and the quality was amazing.
Oh that is such a fun makeup bag! I can see why you had to have it Kim, and it's great you finally got it in the end - even better that they are making it available again, even if it's not as good value for money.
Oh that bag is so cute!
This makeup bag is just darling and totally makes me smile as well, Kim! It's not really something I would gravitate towards, but the cute details are just too cute deny!
This cute makeup bag is adorable.
That is such a great deal! And for 750 points, you are so lucky to get the deal, I never managed to score anything off Sephora rewards because they are out of stock so quick!
I can almost feel the jelly bag! It happens too often I've bought the product because I liked the gwp!
The bag is so cute! I saw the Birthday offer in an email last week! I think its a pretty good deal!
This makeup bag is the cutest! Love it.
Omg, how cute is this makeup bag!? I'm glad you were able to get your hands on it, I can see why it sold out so quickly! Especially since all things 80s and 90s are trending huge this season and this jelly design is bringing flashbacks from my childhood, haha! Thanks for sharing, gorgeous, and I hope you're having the best start to your week so far!
That's such a cute and pretty makeup bag! I'm glad you were able to get your hands on it Kim! That's awesome Sephora had it as one of their rewards!
This is all adorable! I love skinnydip as well and this is def the perfect collaboration!
What a fun bag! Who doesn't love sparkles?
You are so cute Kim! I don't blame you, that is the cutest bag ever! I love using these bags especially when I'm travelling.
I've never heard of Skinny Dip London until I see their collaboration, I had a look on their website and just as you said Kim, their stuff is soooo adorable! Glad that they do international shipping, it seems it's time I fill my cart.
The bag is fun and adorable. I have quit a few makeup bags and have products in each one. I guess in a bid to hide them from my daughter haha.
This bag is adorable!!! And it came with such great goodies. So awesome that you got it. Have you seen the new Clover palettes that are going to be released soon? What do you think of them? I cannot wait to read your review if you purchase it. The colours look pretty light so I don't know if they will suite all skin tones.


In the last part, I discussed some of the challenges that come up both as a student and as a Muslimah attending a Western university. Here are some more issues I have dealt with personally.
Attending school in the West inevitably means attending a co-ed institution, most of the time. Unless you are lucky to go to an all-girls' college (some do exist here!), you will be running into men, whether you mean to or not. First of all, it is allowed to be in a mixed gender environment for the purpose of education. However, this does NOT mean it is okay to flirt and chat with other guys, even if you are in an educational environment. The same rules in dealing with the opposite gender apply here, as well. Which leads to the unavoidable, completely awkward situation of moving your hand away from a guy offering his hand when he meets you. The awkwardness is unavoidable—a solution is to pretend to sneeze into your own hand ;) Joking aside, though…if you take the time to explain to a guy that your religion protects women and that touching is not allowed, he will accept it and respect you. What's better is that he will also keep his distance—a safe but still friendly distance. In the worst-case scenario, the guy will just avoid you at all costs. But these restrictions in Islam are for the best, and Allah subhaana wa ta'aala knows what is best for us.
2) Time Management: It's due TODAY?!
No matter what faith you practice, time management is something that becomes an issue for all college students. In Islam, time is a precious commodity that we as Muslims are blessed with. Time is a means that should be cherished and used wisely. In school, this is especially important: make timelines, write down what is due weekly, and set aside study hours to keep you in check. No one enjoys the feeling of procrastination or having to finish papers at the last minute. So why waste the valuable time you have at school when you can be getting things done? One more note I have learned since attending college is that staying awake after Fajr prayer (Dawn prayer) gives you the chance to get SO much done. There is barakah in that post-sunrise time that's not found throughout the day again. So make use of the time that you are blessed with, and don't let the procrastination lifestyle get to you!
College students are notorious for terrible eating habits: chips, candy bars, pre-packaged foods like Ramen, microwavable dishes, or take out. Constant studying, extracurricular activities, and college life in general takes up so much time that food gets pushed to the back of one's mind. The good news is that it does not have to be that way! The best solution is to make food at home or in your apartment. Sandwiches, salads, and soups, for example, are excellent for lunches. If you live in a dorm, try narrowing down your food list: keep away from oily, heavy, greasy food. Choose fresher, greener, and more colorful food. Vegetables and fruits should be a priority, as well as whole grains instead of white products. Sugar is okay once in a while, like a cupcake here and there. The trick is not to overdo it. As it goes in the Sunnah: 1/3 for food, 1/3 for water, and 1/3 for air with respect to the stomach. Eat enough for your fill, and choose those food that will fuel you longer: fibrous, colorful, and fresh.
With these tips in mind, I pray that all college Muslimahs have a productive year and experience.


text:
UC Irvine, Libraries, University Archives > University of California, Irvine Communications Photographs, Slides Series > Professor Monte Buchsbaum, B.I.C., with cyclotron.
Image / Professor Monte Buchsbaum, B.I.C., with cyclotron.
Title: Professor Monte Buchsbaum, B.I.C., with cyclotron. Date: 1990-02 Collection: University of California, Irvine Communications Photographs, Slides Series Owning Institution: UC Irvine, Libraries, University Archives Source: Calisphere Date of access: January 21 2020 07:52 Permalink: https://calisphere.org/item/c4c641ab-c108-446e-9d71-1c5bffa23046/
UC Irvine, Libraries, University Archives
spcoll@uci.edu
http://special.lib.uci.edu/collections/university-archives/
RE: Calisphere: Request high-resolution copy of item for Professor Monte Buchsbaum, B.I.C., with cyclotron.
Do you need a bigger file? The file available on this site is 667 x 1024 px. To obtain an alternate file type or higher resolution copy, please contact the owning institution.
Item Information.
Professor Monte Buchsbaum, B.I.C., with cyclotron.
University of California, Irvine Libraries. Special Collections and Archives. Irvine, California 92623-9557. http://special.lib.uci.edu. email: spcoll@uci.edu
University of California, Irvine Communications Photographs, Slides Series
This material is provided for private study, scholarship, or research. Property rights reside with the University of California. Copyrights are generally retained by the creators of the records and their heirs, unless transferred to the University of California. It is the responsibility of the researcher to determine who holds the copyright and pursue the copyright owner or his or her heir for permission to publish where the UC Regents do not hold the copyright. For information on use, copyright, and attribution, please visit: http://special.lib.uci.edu/using/publishing.html
AS-061_S00924_002.tif
35mm Slide
University Communications Photographs. AS-061. University of California, Irvine Libraries. Special Collections and Archives. Finding aid for entire collection available at: http://www.oac.cdlib.org/findaid/ark:/13030/c8tt4pp0/
The original slides were collected or created by the UCI Strategic Communications Office. They were transferred to the UCI Libraries University Archives in multiple installments. The slides were scanned by the UCI Libraries in 2015.
: Professor Monte Buchsbaum, B.I.C., with cyclotron.
Institution: UC Irvine, Libraries, University Archives

redpajama_set_name: RedPajamaCommonCrawl


text:
Great South West of Western Australia
Pacific Coast Way Road Trip
SA Road Trip
The Best of the West
The Great Australian Outback Adventure
The Pacific Coast Drive
4WDing
Kids Facilities
Lake or River
Snow Regions
Benalla Tourist Park
| Holiday & Caravan Park
Benalla is on the Broken River, in Victoria's picturesque High Country. Only two hours' drive from Melbourne, Benalla is a great place to spend a weekend in the country and soak up the sites of this interesting region. There's plenty of accommodation on offer at Benalla Tourist Park, whether you're in a caravan, camper van or staying in a cabin. Located in Victoria's picturesque High Country, only two hours' drive from Melbourne, Benalla Tourist Park is a great place to spend a weekend and relax.
BBQ;
Camp Kitchen;
Swimming Pool;
Laundry;
Benalla in Victoria's High Country, just 2 hours from Melbourne, is a beautiful place for a holiday of any length.
The Broken River winds its way through town, surrounded by scenic Botanical Gardens with biking and walking tracks. There's a vibrant art scene in Benalla so you can spot many sculptures and wall art around town on your way to visit the Benalla Art Gallery. For history buffs, this is Ned Kelly country, home to the Stringybark Creek Site. You're only 20 minutes away from the Ned Kelly Museum and Homestead and 30 minutes from the Winton Wetlands.
Benalla Tourist Park, located close to the centre of town, offers comfortable accommodation in this picturesque region, with facilities including a swimming pool, BBQ, picnic area, new camp kitchen and a shared TV room.
On offer is a selection of motel rooms, cabins and villas all with air-conditioning. The cabins and villas are self-contained, including bathroom, kitchenette, lounge, TV and dining area. The motel rooms are studio rooms with bar fridge and bathroom. Towels and linen are provided in all accommodation.
If you are travelling with a caravan, motorhome or tent, the roomy powered grass sites, slab sites, and unpowered sites will suit. Dogs are welcome on all camping sites too!
115 Sydney Road, Benalla, VIC 3672
Contact Park
Kids (3-16yrs)
Infants (0-2yrs)
Cabins only Sites only
G'day Parks
Park Categories
Cabin Guide
Park Login
G'day Parks is Australia's leading community of independent caravan and holiday parks. Formerly known as Top Parks, the G'day Parks network has grown to over 300 parks, making it the largest of its kind in Australia. With more holiday parks in more places, we deliver unmatched choice and freedom, great accommodation and, of course, a warm and welcoming g'day.
Owned by the G'day Group and headquartered in Adelaide, South Australia, G'day Parks invites you to join our tribe and make holiday memories that put a smile in every day.
Stay up to date with park news, specials and events
© G'day Parks 2022
Before you continue...
This website collects data as you browse. As a part of our commitment to ensuring your comfort level with this arrangement, please indicate your preferences.
You can change these settings at any time from our privacy policy page.
Edit my preferences
Essential Required for the website to function
Marketing Measure the success of marketing campaigns
Performance Track page load times and monitor slow areas
Accept all & continue Save


text:
Reassessment of Carrel's Immortal Tissue Culture Experiments
In the 1910s, Alexis Carrel, a French surgeon and biologist, concluded that cells are intrinsically immortal. His claim was based on chick-heart tissue cultures in his laboratory that seemed to be able to proliferate forever. Carrel's ideas about cellular immortality convinced his many contemporaries that cells could be maintained indefinitely. In the 1960s, however, Carrel's thesis about cell immortality was put into question by the discovery that human diploid cells can only proliferate for a finite period.
Subject: Processes, Theories
The Neuron Doctrine (1860-1895)
The neuron doctrine is a concept formed during the turn of the twentieth century that describes the properties of neurons, the specialized cells that compose the nervous system. The neuron doctrine was one of two major theories on the composition of the nervous system at the time. Advocates of the neuron doctrine claimed that the nervous system was composed of discrete cellular units. Proponents of the alternative reticular theory, on the other hand, argued that the entire nervous system was a continuous network of cells, without gaps or synapses between the cells.
The Hayflick Limit
The Hayflick Limit is a concept that helps to explain the
mechanisms behind cellular aging. The concept states that a normal human
cell can only replicate and divide forty to sixty times before it
cannot divide anymore, and will break down by programmed cell death
or apoptosis. The concept of the Hayflick Limit revised Alexis
Carrel's earlier theory, which stated that cells can replicate
themselves infinitely. Leonard Hayflick developed the concept while
at the Wistar Institute in Philadelphia,
The Formation of Reticular Theory
In the nineteenth century, reticular theory aimed to describe the properties of neurons, the specialized cells which make up the nervous system, but was later disconfirmed by evidence. Reticular theory stated that the nervous system was composed of a continuous network of specialized cells without gaps (synapses), and was first proposed by researcher Joseph von Gerlach in Germany in 1871.
Frederik Ruysch, working in the Netherlands, introduced the term epithelia in the third volume of his Thesaurus Anatomicus in 1703. Ruysch created the term from the Greek epi, which means on top of, and thele, which means nipple, to describe the type of tissue he found when dissecting the lip of a cadaver. In the mid nineteenth century, anatomist Albrecht von Haller adopted the word epithelium, designating Ruysch's original terminology as the plural version. In modern science, epithelium is a type of animal tissue in which cells are packed into neatly arranged sheets.
Subject: Theories, Processes
David Reimer and John Money Gender Reassignment Controversy: The John/Joan Case
In the mid-1960s, psychologist John Money encouraged the gender reassignment of David Reimer, who was born a biological male but suffered irreparable damage to his penis as an infant. Born in 1965 as Bruce Reimer, his penis was irreparably damaged during infancy due to a failed circumcision. After encouragement from Money, Reimer's parents decided to raise Reimer as a girl. Reimer underwent surgery as an infant to construct rudimentary female genitals, and was given female hormones during puberty.
Charles Darwin's Theory of Pangenesis
In 1868 in England, Charles Darwin proposed his pangenesis theory to describe the units of inheritance between parents and offspring and the processes by which those units control development in offspring. Darwin coined the concept of gemmules, which he said referred to hypothesized minute particles of inheritance thrown off by all cells of the body. The theory suggested that an organism's environment could modify the gemmules in any parts of the body, and that these modified gemmules would congregate in the reproductive organs of parents to be passed on to their offspring.
Germ Layers
A germ layer is a group of cells in an embryo that interact with each other as the embryo develops and contribute to the formation of all organs and tissues. All animals, except perhaps sponges, form two or three germ layers. The germ layers develop early in embryonic life, through the process of gastrulation. During gastrulation, a hollow cluster of cells called a blastula reorganizes into two primary germ layers: an inner layer, called endoderm, and an outer layer, called ectoderm.
The Germ-Plasm: a Theory of Heredity (1893), by August Weismann
Friedrich Leopold August Weismann published Das
Keimplasma: eine Theorie der Vererbung (The Germ-Plasm: a
Theory of Heredity, hereafter The Germ-Plasm) while
working at the University of Freiburg in Freiburg, Germany in 1892.
William N. Parker, a professor in the University College of South
Wales and Monmouthshire in Cardiff, UK, translated The
Germ-Plasm into English in 1893. In The Germ-Plasm,
Weismann proposed a theory of heredity based on the concept of the
Edward Drinker Cope's Law of Acceleration of Growth
The Law of Acceleration of Growth is a theory proposed by Edward Drinker Cope in the US during the nineteenth century. Cope developed it in an attempt to explain the evolution of genera by appealing to changes in the developmental timelines of organisms. Cope proposed this law as an additional theory to natural selection.
The Source-Sink Model
The source-sink model, first proposed by biologist Francis Crick in 1970, is a theoretical system for how morphogens distribute themselves across small fields of early embryonic cells. A morphogen is a substance that determines the fate and phenotype of a group of cells through a concentration gradient of itself across that group. Crick's theory has been experimentally confirmed with several morphogens, most notably with the protein bicoid , the first discovered morphogen. The model provides a theoretical structure for the understanding of some features of early embryonic development.
Trial of Madame Restell (Ann Lohman) for Abortion (1841)
In the spring of 1841, abortionist Ann Lohman, called Madame Restell, was convicted for crimes against one of her abortion clients, Maria Purdy. In a deathbed confession, Purdy admitted that she had received an abortion provided by Madame Restell, and she further claimed that the tuberculosis that she was dying from was a result of her abortion. Restell was charged with administering an illegal abortion in New York and her legal battles were heavily documented in the news.
Endothelium
The endothelium is the layer of cells lining the blood vessels in animals. It weighs more than one kilogram in adult humans, and it covers a surface area of 4000 to 7000 square meters. The endothelium is the cellular interface between the circulating blood and underlying tissue. As the medium between these two sets of tissues, endothelium is part of many normal and disease processes throughout the body.
Human Papillomavirus (HPV) Strains 16 and 18
The Human Papillomavirus (HPV) strains 16 and 18 are the two most common HPV strains that lead to cases of genital cancer. HPV is the most commonly sexually transmitted disease, resulting in more than fourteen million cases per year in the United States alone. When left untreated, HPV leads to high risks of cervical, vaginal, vulvar, anal, and penile cancers. In 1983 and 1984 in Germany, physician Harald zur Hausen found that two HPV strains, HPV-16 and HPV-18, caused cervical cancer in women. In the early twenty first century, pharmaceutical companies Merck & Co.
The Debate over DNA Replication Before the Meselson-Stahl Experiment (1953–1957)
Between 1953 and 1957, before the Meselson-Stahl experiment verified semi-conservative replication of DNA, scientists debated how DNA replicated. In 1953, James Watson and Francis Crick proposed that DNA was composed of two helical strands that wound together in a coil. Their model suggested a replication mechanism, later termed semi-conservative replication, in which parental DNA strands separated and served as templates for the replication of new daughter strands.


It's not easy being an investor in some merger-and-acquisition deals.
Take Monday's announcement of a big—$34 billion, plus $9 billion in debt—acquisition by Fidelity National Information Services (ticker: FIS), or FIS, of Worldpay (WP). Both are analyst picks; neither are household names. FIS is a $9.2 billion in revenue broad-based provider of digital-processing services between retail and financial services, including banks. Worldpay is a $3.9 billion major processor of merchant transactions—mobile, online, and in-store, with an emphasis on e-commerce.
Both have emerged from a scrum of M&A transactions over the years. The payments-processing industry, which is layered, fragmented and global, has seen rapid growth and consolidation: In January Fiserv (FISV) said it was buying First Data (FDC) for $22 billion; Worldpay itself was acquired by Vantiv for $10.4 billion last year and took its name; and FIS has grown for years through M&A, including buying Sungard for $9 billion in 2015. Two things stand out: There's a Mixmaster full of M&A here, intercut with private equity ownership.
FIS and Worldpay presented this deal, with its modest 14% premium, as a consolidation play. Worldpay will provide FIS another service to sell to financial services customers with its big global footprint. "Scale matters in our rapidly changing industry," the press release quotes FIS Chairman and CEO Gary Norcross.
In fact, payments processing is a high-tech, fast-changing business that requires economies of scale to make money—huge commodity-like volumes—and generates lots of valuable data. There are regularly new entrants with new technologies. Security is important. And both the technology and the business are complex and goal oriented: fully integrated, global, frictionless, end-to-end processing for mostly retail payments. Those conditions—scale, integration, disruption, data—catalyze M&A. Think Cisco Systems (CSCO) and its M&A program.
By most measures, FIS has done a good job on both operations and M&A: Earnings have been growing, it's kept expenses down, and the company has successfully integrated acquisitions. True, its trailing-12-month return on equity of 16.7% lags the industry's 20.3%. And shares have advanced 8.7% in the past year, compared with 12.8% for the industry, according to Zacks. FIS promises it will extract $400 million a year in synergies from the Worldpay deal, and boost organic growth to 9% from 6%. And the combined company will produce $15 billion in revenue and $7 billion in earnings before interest, taxes, depreciation, and amortization in three years, which would reduce the debt on the deal.
This deal may, in fact, work out; a number of analysts like it. But if you didn't already know, would you guess what Worldpay actually does—and what it's No. 1 at? Quick, parse the difference between fintech, e-commerce, global payments, and merchant payments? How do you judge solutions sets, global communities and client focus? Define personalization. And then there's the accelerating future! Either Norcross is trying to obscure how these two companies fit together in a complicated jigsaw puzzle, or this is just how payment-processing folks talk. And this isn't just FIS; it's common on many M&A calls, which are essentially pitch sessions to investors, with giddy analysts offering congrats. It makes for a deal that's tough to untangle, not to say predict.
Does it matter? In any given M&A deal, there are a handful of well-publicized ways deals can succeed, and a thousand obscure ways they can fray. Bankers, analysts, and management tend to focus on often-simple upsides: synergies, new products and talent, scale and scope, pricing power and leverage. But failure? Talent flees, products fizzle, synergies evaporate, lawsuits accumulate, innovation peels away like old wallpaper, and new deals always beckon.
That may not be the case here. But it's not a huge surprise that FIS stock fell by 5% on Monday after opening at $112, before closing 2% down, while Worldpay soared 9.3%. If that continues, FIS might have to hike the price or fend off another bidder—perhaps one that can really explain why it's such a good deal in the first place.


BSNL Punjab is offering extra talktime of Rs. 150 on RC/Top-up 1100. The usual talktime offered on RC/Top-up 1100 is 1050 with 365 days of validity but now subscribers will get a talktime of Rs. 1200 but a reduced validity of 270 days. The offer is not applicable on GSM PCO and can be availed from 15-10-2011 to 13-11-2011.


baby girl pram new baby card by jenny arnott cards gifts .
personalised baby girl congratulations card by philly brit .
free printable baby cards my free printable cards com free .
baby girl in stroller thank you notes personalized note cards .
congratulations greeting card baby girl lils cards .
new baby girl lullaby cd greeting card cards .
little baby girl cards molly mae new baby cards new baby girl cards .
personalised new baby girl card design 4 .
baby girl dress advice cards paperstyle .
handmade card congratulations to new baby girl luv the die .
baby girl card buzz ideazz .
baby girl card oliver bonas .
new baby girl cards funky pigeon .
new baby girl paper salad greeting card cards .
cards for new baby girl rome fontanacountryinn com .
cute baby girl card vector free download .
baby girl birth announcement cards set of 8 cards mini me gift .
paper cuts by lisa baby girl onsie card online store powered by .
illustrtion of baby girl card birthday card royalty free cliparts .
amazon com 30 baby shower invitations girl with envelopes 30 pack .
precious baby girl new baby greeting card cards .
baby pregnancy announcement cards funky pigeon .
10 personalised baby girl birth announcement thank you photo cards n102 .
welcome little one new baby girl card caroline gardner uk .
baby cards baby girl cards new baby cards new baby girl card .
1st birthday card baby girl meylah .
age 1 girl birthday card baby girl teddy bear pink polka dots .
congratulations on your new baby girl card dot2dot cards gifts .
precious baby girl free baby shower new baby card greetings island .
pearly pink pram new baby girl card jt cards .
personalised baby girl letter washing line card .
personalised new baby girl card baby grow handmade ebay .
tweet little baby girl new baby girl card 2 80 a great range of .
pink flowers star baby girl congrats send this greeting card .
baby girl arrival card stock vector illustration of celebrate .
baby girl card etsy .
milestone cards watercolour baby girl audrey me .
baby girl card by neetika agarwal dribbble dribbble .
pink a new arrival card baby girl card karenza paperie .
baby girl congratulations cards under fontanacountryinn com .
pink greeting card with teddy bear for baby girl stock vector .
baby girl milestone cards milestone cards tiny viking co .
baby girl greeting card .
baby girl greeting card illustration .
new baby girl verse nbgv002 card sentiments baby girl cards .
baby girl card tree design .
1st birthday personalised card for girls by tandem green .
welcome baby girl card vector free download .
baby girl milestone cards rachelle rachelle .
baby girls first birthday card karenza paperie .
baby girl card stock vector illustration of cartoon 21941888 .
birth announcement cards baby announcement cards .
baby girl card crib design .
handmade baby girl greeting card youtube .
birth announcements and baby thank you photo cards for girls with bow .
cards baby girl rome fontanacountryinn com .
bundle of joy baby girl and baby boy cards pictures photos and .
congratulations cards for all occasions baby congratulations .
free printable baby cards gallery 2 .
second nature new baby girl keepsake card cards .
amazon com books for baby request cards girl baby shower .
how to make a welcome baby girl card .
baby girl card under fontanacountryinn com .
a new baby girl congratulations card .
printable baby shower congratulations cards image cabinets and .
birth announcements and baby thank you photo cards for girls with .
baby girl birth greeting card royalty free vector image .
baby girl milestone cards .
119 best baby cards images handmade cards kids cards new .
babys first birthday card personalised first birthday card .
baby girl shower card stock vector illustration of congratulation .
baby girl pink butterfly printable baby shower wishes for baby .
baby girl greeting card by allison black lagom design .
baby shower invitation boys girls birthday greeting card gifts baby .
cute pink owl girl baby shower invitation card .
10 piece lot baby shower invitation cards baby boy baby girl shower .
baby custom card baby boy or girl neutral rainbow buttons newborn .
baby milestone cards uk girl baby shower gifts girl new baby girl .
new baby girl greeting card by hanna melin notonthehighstreet com .
babys 1st birthday thank you note cards baby girl mandys moon .
baby cards ireland bunting baby card .
baby girl thank you cards planet cards co uk .
birthday card for baby girl stock vector illustration of gift .
baby girl greeting card barca fontanacountryinn com .
handmade new baby girl card flower 620944688494 ebay .
congratulations on your new wee bairn card new baby girl card .
full of giggles new baby girl card greeting cards hallmark .
pram shaped card baby girl cup612147_1028 craftsuprint .
new granddaughter cards make it special funky pigeon .
baby girl arrival announcement card stock vector illustration of .


text:
Lost Highway bezeichnet:
 Lost Highway (Album), ein Album der Rockband Bon Jovi
 Lost Highway, einen Film des Regisseurs David Lynch aus dem Jahr 1996
 Lost Highway (Oper), eine Oper von Olga Neuwirth und Elfriede Jelinek, basierend auf dem Film von David Lynch
 Lost Highway (Label), ein Unterlabel der Plattenfirma Universal Music
 einen Song von Leon Payne, dessen bekannteste Interpretation von Hank Williams stammt

redpajama_set_name: RedPajamaWikipedia


text:
AN ACT GRANTING LYNDA FELICIANO A FRANCHISE TO CONSTRUCT, OPERATE AND MAINTAIN AN ICE PLANT AND COLD STORAGE IN THE MUNICIPALITY OF MAKATI, PROVINCE OF RIZAL.
Licenses -- Ice plant and cold storage
Lynda Feleciano -- Licenses
Licenses -- Lynda Feleciano
Official Gazette vol. 66 no. 9 page 2039 (3/2/1970)
H. No. 17370 / 66 OG No. 9, 2039 (March 2, 1970)
Be it enacted by the Senate and House of Representatives of the Philippines in Congress assembled:SECTION 1. Subject to the terms and conditions established in this Act, in Commonwealth Act Numbered One hundred forty-six, as amended, and to the provisions of the Constitution, there is granted to Lynda Feliciano, for a period of twenty-five years from the approval of this Act, the right, privilege and authority to construct, operate and maintain an ice plant and cold storage in the Municipality of Makati, Province of Rizal, for the purpose of manufacturing ice for distribution and sale, and for supplying cold storage in the said municipality and to charge and collect a schedule of prices and rates therefor which schedule shall at all times be subject to regulation by the Public Service Commission or its legal successor.SEC. 2. The grantee shall manufacture and supply ice up to the limit of the capacity of the plant, said limit to be determined by the Public Service Commission, and sell the same at rates to be determined by the Public Service Commission.SEC. 3. All the apparatus and appurtenances to be used by the grantee shall be modern, safe and first class in every aspect, and the grantee shall, whenever the Public ServiceCommission shall determine that public interest reasonably requires it, change or alter any of its apparatus and appurtenances at grantee's expense.SEC. 4. If the grantee shall not commence the construction of the ice plant and cold storage within two years from the approval of this Act, unless prevented by an act of God or force majeure, martial law, riot, civil commotion, usurpation by a military power or any other cause beyond the grantee's control, this franchise shall become null and void.SEC. 5. This franchise is granted with the understanding and upon the conditions that its terms shall not exceed twenty-five years, that it shall be subject to amendment, alteration, or repeal by the Congress of the Philippines when the public interest so requires, and that the grantee shall be subject to all existing taxes under the National Internal Revenue Code.SEC. 6. The books, records and accounts of the grantee shall always be open to the inspection of the Auditor General or his authorized representatives, and it shall be the duty of the grantee to submit to the Auditor General quarterly reports in duplicate showing the gross receipts for the quarter past.SEC. 7. In the event of any competing person, natural or juridical, receiving from the Congress of the Philippines a j similar franchise in which there shall be any term or terms more favorable than those herein granted or tending to place the herein grantee at any disadvantage, then such, term or terms shall ipso facto become a part of the terms hereof and shall operate equally in favor of the grantee as in the case of such competing person.SEC. 8. In consideration of this franchise the grantee shall pay a tax equivalent to five percent of her gross income.SEC. 9. This Act shall take effect upon its approval.Enacted without executive approval, June 21, 1969.


Once you've raked the leaves and needles, it's tempting to stop thinking about tree health.
This is a great time, though, to do a few things for your trees. Here are some tasks that should be completed over the next couple of months.
Pruning: Deciduous trees are best pruned during their dormant period. Check all trees for signs of diseased, damaged or broken branches. Remove any you find. Look for branches with brown leaves that persist on the stem, on the most recent growth. You can tackle two phases of treatment during late fall and winter. First, prune off the diseased branches. During the growing season, you must be cautious, cleaning the pruning blade between each cut. But in the dormant season, the bacteria are wintering over in the cankers you find in diseased branches. Cross-contamination is not as likely because the bacteria are inactive. Remove the affected branches below the damage, such as cankers. The second step is to contact an arborist and schedule a bactericide spray during the bloom season. A good arborist will show up at just the right time to apply the antibiotic, when the tree is in bloom. The antibiotic will kill the bacteria, and rain will wash them away. For more information on fire-blight management, consult the Cornell University fact sheet at nysipm.cornell.edu/factsheets/treefruit/diseases/fb/fb.asp.


I saw this on Fuzzy Little Balrog and I thought it was a lovely idea to share a bit about my relationship.
All is never black and white and I think its great to give a bit of an insight into my life.
We worked together at a make over studio. I'd been there 6 months or so and he was the new guy that was working in another department. I noticed him straight away and once we got talking we clicked immediately. I've never had such a strong connection with a person before.
Sam was in the middle of training with another member of staff. She was introducing him to their clients as "Paul" for some reason. He was stood next to me and lent over and went "my name isn't even Paul" I giggled and he went off to see to the clients. I don't really remember the first proper conversation but your departments worked closely together so I had to spend a lot of time alongside him which was great.
We will be celebrating 4 years together on May 11th when we go to Disneyland!
WELL. The first was a big of a naughty drunken kiss when we both had partners and celebrated our joint birthdays out with work. It was over pretty quickly as we knew it really shouldn't have happened. Just over 2 years later it happened again in the form of Sam catching me off guard at the station after leaving work together. There had been a power cut at work and we were sent home early. There had been a lot of flirting in the day, lots of being silly about it being dark at work and everything being a bit more exciting. He turned to me and said what would you do if I kissed you right now (smooth, I know) and then kissed me before I got on my train and went home feeling really happy but with my head all over the place as things were finally going somewhere after 2 years.
What is your favourite quality of his?
His passion. Anything he does he really throws himself into it. He gives his all with everything and is just amazing.
What is your favourite physical feature?
His arms, he's really strong so he has great arms. He has a very pretty face too. I an see him looking like Hugh Jackman when he gets older which is a good thing!
What would be your ideal trip away as a couple?
I can't wait to go to Disneyland in May, it's going to be lovely to see Mason's reaction to everything. But for us as a couple I think the best type of holiday for us would be something really luxurious and flashy. Somewhere hot and exotic where we can relax, have massages and eat amazing food.
Have you met his family?
I have. His Granny is hilarious and I love her. He's got loads of brothers so it's hard to keep track of them all but I have huge respect for his bother Jordan, he's a good egg.
Do you have plans together?
Of course. We have a child together so it's kind of hard not to. In a nutshell it's just to be happy and continue to do exciting things together.
What is your favourite memory together?
That is a really tough one. I have so many from when we were friends or from the early stage of being together. We used to just sit in his room chatting for hours and hours. We had next to no money and would have sandwiches for dinner. I remember thinking 'If Im this happy now when we have nothing, imagine how perfect things are gong to be in the future". But Mason being born was obviously the biggest and most amazing memory, Sam completely broke down and I've never seen him so loved and proud of me.
If you fill this in too please let me know in the comments, I love reading this sort of thing!


text:
<?php
namespace phpbu\Laravel\Cmd;

class PrinterTest extends \PHPUnit\Framework\TestCase
{
    /**
     * Tests Printer::write
     */
    public function testWrite()
    {
        $command = $this->getMockBuilder('\\phpbu\\Laravel\\Cmd\\Backup')
                        ->disableOriginalConstructor()
                        ->getMock();
        $command->expects($this->once())
                ->method('info');

        $printer = new Printer($command);
        $printer->write('foo');
    }

    /**
     * Tests Printer::writeWithColor
     */
    public function testWriteColored()
    {
        $command = $this->getMockBuilder('\\phpbu\\Laravel\\Cmd\\Backup')
                        ->disableOriginalConstructor()
                        ->getMock();
        $command->expects($this->once())
                ->method('info');

        // create printer with active debug
        $printer = new Printer($command, false, true);

        // create backup failed event mock
        $event = $this->getMockBuilder('\\phpbu\\App\\Event\\Backup\\Failed')
                      ->disableOriginalConstructor()
                      ->getMock();

        // call event handler to trigger colored write
        $printer->onBackupFailed($event);

        $this->assertTrue(true);
    }
}

redpajama_set_name: RedPajamaGithub


text:
Q: Usage of bodyParser.text() in Express with EJS I am building a search engine app with Express that queries the Twitter API v1.1. Currently, I am trying to submit the search string to my server by parsing form data with the bodyParser module.  Here is the code:
index.ejs
...
<form method="GET" action="/results">
    <input id="input" type="text" name="search">
    <button id="searchButton">+</button>
</form>
...

server.js
var express = require('express'),
    bodyParser = require('body-parser');

var app = express();

var port = process.env.PORT || 8080;

app.set('view engine', 'ejs');

app.use(express.static(__dirname + "/public");

var urlencodedParser = bodyParser.urlencoded({ extended: false })

app.get('/', function(req, res) {
    res.render('index');
});

app.get('/results', urlencodedParser, function (req, res) {
    console.log(req.body);
    res.render('results')
});

app.listen(port, function() {
    console.log('Our app is running on http://localhost:' + port);
});

The code shown will return { } to the console.  If I try and access req.body.search it returns undefined (obviously).  What is the issue here?  Why is it not logging my search string?

A: You're using the wrong body decoder. If you're submitting a form (application/x-www-form-urlencoded is the default enctype for forms), you will need bodyParser.urlencoded() instead of bodyParser.text(). The latter is for plaintext request data, not form data.
Additionally, you should also be using method="POST" and a POST route (app.post('/results', ...)) instead of method="GET" and a GET route. Since GET requests almost never have a body, the browser instead transforms the form data into a querystring that is then appended to the url itself. This means your form data is currently in req.query instead of req.body. Switching to POST though will cause the browser to send the form data in the request body, and the form data will be in req.body as expected.

redpajama_set_name: RedPajamaStackExchange