text:
\section{$\beta$-triads and a Structural Theorem}
	\label{43}
	\input{triad}

	\section{Improved Bounds on TSPN}
	\label{44}
	\input{bound}

	\section{The Fermat-Weber Approach}
	\label{45}
	\input{fermat}

	\subparagraph*{Acknowledgements.}

	The author would like to thank Prof. Samir Khuller for suggesting the problem and for inspiring conversations on the topic.

	\small
	\bibliographystyle{abbrv}

\subsection{Disjoint Uniform Disks}
We will now show the proof of Theorem~\ref{thm:mama}.
\begin{proof} Assume the  $TSPN^*$ is not a straight line. We start by singling out the $\beta$-triads and considering the two orderings $\sigma$ and $\sigma'$ from Theorem ~\ref{thm:triadbound}. If there are $k_1$ $\beta$-triads $\mathcal{T}_1, \ldots, \mathcal{T}_{k_1}$ spanning edges of total length $L_T$, we get that:
  \begin{align*}
   |TSP^*| &\leq \displaystyle{\frac{|TSP(\sigma)| + |TSP(\sigma')|}{2}} \\
            &\leq \displaystyle{ |TSP(\sigma \cap \sigma')| + L_T + 3\sqrt{3}R \cdot k_1}.
  \end{align*}

Observe that $TSPN(\sigma \cap \sigma')$ is a collection of disjoint paths. From all of these paths, we further extract each from these a total of $k_2$ subpaths $\mathcal{G}_1,\ldots, \mathcal{G}_{k_2}$ consisting of good edges. Notice that the remaining subpaths left in $\sigma \cap \sigma'$ consist of bad edges which do not form a $\beta$-triad.  Suppose we obtain $l$ such  remaining subpaths paths $\mathcal{B}_1,\ldots,  \mathcal{B}_l$. In other words, we have decomposed the TSPN into three categories of subpaths:
 \begin{itemize}
  \item $k_1$ $\beta$-triads  $\mathcal{T}_1, \ldots, \mathcal{T}_{k_1}$,
  \item $k_2$ paths $\mathcal{G}_1,\ldots, \mathcal{G}_{k_2}$ that cover the remaining good edges, and
  \item $l$ paths  $\mathcal{B}_1,\ldots,  \mathcal{B}_l$ that consist only of bad edges which do not form $\beta$-triads.
 \end{itemize}

We are now ready to evaluate the detour that each of these paths takes. For each $i \in [1,k_2]$ let $\psi_i$ the natural order on the disks associated with $\mathcal{G}_i$ and let $n_i$ be the number of edges in $\mathcal{G}_i$. We have that:
\begin{center}
 $|TSP(\psi_i)| \leq |TSPN(\psi_i)| + ( 1+\cos \beta)R \cdot n_i$.
\end{center}

When it comes to the paths $\mathcal{B}_j$, with $j \in [1,l]$, let $\sigma_j$ be their natural associated orders and let $m_j$ be the number of edges it contains. We have that $|TSP (\sigma_j)| \leq |TSPN(\sigma_j)| + 2R \cdot m_j$.

 Let $N = \sum_{i=1}^{k_2} n_i$ be the total number of edges in $\mathcal{G}_1,\ldots, \mathcal{G}_{k_2}$ and $M = \sum_{j=1}^{l} m_j$ the total number of edges in $\mathcal{B}_1,\ldots,  \mathcal{B}_l$. By construction, we decomposed $TSPN(\sigma \cap \sigma')$ into these two groups of edge disjoint paths and we therefore get that:
 \begin{align*}
   |TSP(\sigma \cap \sigma')| &= \displaystyle{\sum_{i=1}^{k_2}} |TSP(\psi_i)|  +  \displaystyle{\sum_{j=1}^{l}} |TSP(\sigma_j)| &&\\
                             &\leq \displaystyle{\sum_{i=1}^{k_2} \Big( |TSPN(\psi_i)| + (1 +\cos \beta) R \cdot n_i  \Big)}\\
                             & +  \displaystyle{\sum_{j=1}^{l} \Big( |TSPN(\sigma_j)| + 2R \cdot m_j \Big)} &&\\
                            &\leq |TSPN(\sigma \cap \sigma')| + (1 +\cos \beta)RN + 2R M. &&
 \end{align*}

 Including the $\beta$-triads back into our bound, we get that:
 \begin{align*}
 |TSP^*| &\leq \frac{|TSP(\sigma)| + |TSP(\sigma')|}{2}&& \\
         & \leq |TSP(\sigma \cap \sigma')| + L_T + 3\sqrt{3}R \cdot k_1 &&\\
         & \leq |TSPN(\sigma \cap \sigma')| + L_T +  3\sqrt{3}R \cdot k_1 +  (1 +\cos \beta)  RN + 2RM && \\
         & \leq |TSPN| + 3\sqrt{3}R \cdot k_1 +  (1 +\cos \beta)RN + 2RM. &&
 \end{align*}

In other words, we've expressed the total detour of the $TSPN$ according to  edges that participate in $\beta$-triads, edges in $\mathcal{G}_1,\ldots, \mathcal{G}_{k_2}$  and edges in  $\mathcal{B}_1,\ldots,  \mathcal{B}_l$. By construction, none of these paths share edges and so $3k_1 + N + M = n$. Let $K = 3k_1 + N$ be the total number of edges either in a $\beta$-triad or in $\mathcal{G}_1, \ldots, \mathcal{G}_{k_2}$ and since $\sqrt{3} \leq 1 +\cos \beta$, we have that:
\begin{center}
 $|TSP^*| \leq |TSPN| + (1 + \cos \beta)R \cdot K + 2R \cdot (n-K)$.
\end{center}

\textbf{Case $1$: when $K \geq \frac{n}{2}$.} In this situation, we have that:
\begin{center}
 $|TSP^*| \leq |TSPN^*| + \displaystyle{ \frac{3 +\cos \beta}{2}  }\cdot R \cdot n$.
\end{center}

The average detour per edge $\frac{3+\cos \beta}{2}$ is better than the $2R$ bound, but it is constrained by the choice of $\beta \in [0, \pi/12]$, which means that the best we could hope for is an average detour of $\frac{1}{2}(3+\cos \frac{\pi}{12})R < 1.983R$. We note that the average detour in the H\"{a}me, Hyyti\"{a} and Hakula conjecture is $\sqrt{3}R \approx 1.732R$. Using Lemma ~\ref{lem:TSPNtour} gives us that
\begin{align*}
|TSP^*| &\leq  \displaystyle{ \Big(1 +  \frac{2}{\pi} \cdot(3+\cos \beta)\Big)} \cdot |TSPN^*| + \\
        &+ 2 \cdot (3+\cos \beta)R&.
\end{align*}
For large $n$, the $1 + \frac{2}{\pi} \cdot(3+\cos \beta)$ term will dominate our approximation factor and is at most $3.525$, when $\beta = \pi/12$.

\textbf{Case $2$: when $K < \frac{n}{2}$.} In this situation, even the overall detour might be large, we will show that in fact, in this case, $TSP^*$ is a $2$-approximation and therefore, the best that it can be in general. We know that each path $\mathcal{B}_j$ consists of bad edges which do not form any $\beta$-triads. In other words, if $P_1P_2$ is an edge in it, then we know that $|O_1O2| > R/\sin(2\beta)$ which in turn means that $|P_1P_2| > (1/\sin(2\beta)-2) \cdot R$. Overall we have that:
\begin{align*}
&&|TSPN| &\geq \displaystyle{\sum_{j=1}^{l} |\mathcal{B}_j|} \geq \Big(\frac{1}{\sin{(2\beta})} -2\Big)R \cdot (n-K) &&\\
&&       &\geq \Big (\frac{1}{2\sin{(2\beta)}} -1 \Big)R \cdot n.&&
\end{align*}

Since the total detour could be at most $2R$ per edge, we get that:
\begin{center}
$|TSP^*| \leq \displaystyle{ \Big( 1 + \frac{2}{\frac{1}{2\sin{(2\beta)}} -1} \Big )} \cdot |TSPN|$.
\end{center}
When $\beta = \frac{1}{2}\arcsin\frac{1}{6}$, the detour from Case $1$ becomes $\frac{3 + \cos \beta}{2} \approx 1.998$ and the approximation factor from Case $2$ becomes exactly $2$.
We note that the machinery described can be used to obtain more nuanced results. In particular, lower choices for $\beta$ will drive the approximation factor in Case $2$ even lower than $2$, at the expense of a higher detour bound for Case $1$.

\end{proof}

More generally, we can consider a parameter $\alpha >1$ that we will set later in the proof. We include here only the aspects that change. Depending on whether $K \leq \frac{n}{\alpha}$ or not, we will employ different lower bounds on $|TSPN|$, in a similar fashion as before.

\textbf{Case $1$: when $K \geq \frac{n}{\alpha}$.} In this situation, we have that:
\begin{center}
 $|TSP^*| \leq |TSPN^*| + \displaystyle{ \frac{1 +\cos \beta + 2(\alpha-1)}{\alpha}  }\cdot R \cdot n$.
\end{center}
 Using Lemma ~\ref{lem:TSPNtour} gives us that
\begin{align*}
|TSP^*| &\leq  \displaystyle{ \Big(1 +  \frac{4}{\pi} \cdot \frac{1 +\cos \beta + 2(\alpha-1)}{\alpha} \Big)} \cdot |TSPN^*| + 4 \cdot \frac{1 +\cos \beta + 2(\alpha-1)}{\alpha}R\\
         &\leq \displaystyle{ \Big(1 +  \frac{4}{\pi} \cdot \frac{1 +\cos \beta + 2(\alpha-1)}{\alpha} \Big)} \cdot |TSPN^*|+ 8R\\
         &\leq \displaystyle{ \Big(1 +  \frac{8}{\pi} - \frac{4}{\pi} \cdot \frac{1 -\cos \beta}{\alpha} \Big)} \cdot |TSPN^*|+ 8R.
\end{align*}

\textbf{Case $2$: when $K < \frac{n}{\alpha}$.}
 We know that each path $\mathcal{B}_j$ consists of bad edges which do not form any $\beta$-triads. In other words, if $P_1P_2$ is an edge in it, then we know that $|O_1O2| > R/\sin(2\beta)$ which in turn means that $|P_1P_2| > (1/\sin(2\beta)-2) \cdot R$. Overall we have that:
\begin{align*}
&&|TSPN| &\geq \displaystyle{\sum_{j=1}^{l} |\mathcal{B}_j|} &&\\
&&       &\geq \Big(\frac{1}{\sin{(2\beta})} -2\Big)R \cdot (n-K) &&\\
&&       &\geq \frac{\alpha-1}{\alpha} \cdot\Big(\frac{1}{\sin{(2\beta})} -2\Big)R \cdot n.&&
\end{align*}
Since the total detour could be at most $2R$ per edge, we get that:
\begin{center}
$|TSP^*| \leq \displaystyle{ \Big( 1 + \frac{\alpha}{\alpha-1} \cdot \frac{2}{\frac{1}{\sin{(2\beta)}}-2} \Big )} \cdot |TSPN|$.
\end{center}

If we want to achieve a factor $2$-approximation in Case $2$, we need to have $\beta \leq \frac{1}{2}\arcsin(\frac{1}{4}) $ and set
\begin{center}
 $\alpha = 1 + 2 / (\frac{1}{\sin{(2\beta)}}-4)$.
\end{center}
In this case, the detour in Case $1$ becomes $2-(1-\cos \beta)(2 - 1/(1-2\sin{(2\beta)})$ which achieves a minimum of $\approx 0.998$ on the interval $[0,\frac{1}{2}\arcsin(\frac{1}{4})]$. Setting $\alpha = 1 + 2/(c/{\sin(2\beta)} - 2c-2)$ for $c = 2.53$ and $\beta = 0.1831$ gives us that both of these cases lead to a $2.53$-approximation.

\subsection{Overlapping Uniform Disks}

We discuss how the analysis from the disjoint case carries over to the case of overlapping disks. As we mentioned before, the best known approximation for this case is by Dumitrescu and T\'{o}th~\cite{dumitrescu2016traveling}. In general, approaches for this case take advantage of known analyses for the disjoint case and adapt them in a smart way to the overlapping case. We begin by roughly describing the technique of Dumitrescu and T\'{o}th~\cite{dumitrescu2016traveling} and the show how the analysis changes when we use our framework.

Specifically, Dumitrescu and T\'{o}th~\cite{dumitrescu2016traveling} start by computing a monotone maximal set of disjoint disks $\mathcal{I}$ by greedily selecting the leftmost disk and deleting all of the other input disks that intersect it. Let $k$ be the size of the set we end up wth. They then compute an approximate TSP tour on the centers of the disks in $\mathcal{I}$, either using the available schemes~\cite{arora1998polynomial,mitchell1999guillotine} or Christofides~\cite{Chr76}. We call this tour $T_\mathcal{I}$. They then augment this tour in such a way that we visit all the input disks, not just the ones in $\mathcal{I}$. Before we discuss the augmentation part, we first define some notation and mention some bounds that follow naturally.

Let the optimal TSP tour on the centers in $\mathcal{I}$ be $TSP^*_\mathcal{I}$. The eventual tour $T_\mathcal{I}$ that we compute will be an $a$-approximation to $TSPN^*_\mathcal{I}$ so we have that:
\begin{equation}\label{bound:1}
 |T_\mathcal{I}| \leq a \cdot |TSP^*_\mathcal{I}|.
\end{equation}

On the other hand, we know that this set of disks also has an associated optimal TSPN tour, which we call $TSPN^*_{I}$. Finally, we denote the optimal TSPN tour on all the disks by $TSPN^*$. We know that the tour on $\mathcal{I}$ is a lower bound:
\begin{equation}\label{bound:2}
 |TSPN^*_\mathcal{I}| \leq |TSPN^*|.
\end{equation}
The size of our final solution will be compared to $|TSPN^*|$ and to that end, we use lower bounds on $|TSPN^*_\mathcal{I}|$ in conjunction with (\ref{bound:2}) to get lower bounds on $|TSPN^*|$. This is the part where our new framework will come in, because $|TSPN^*_\mathcal{I}|$ is a tour on disjoint disks by definition.

The next step is to augment $T_\mathcal{I}$ with detours of length $O(R)$ along the disks in $\mathcal{I}$ such that it touches every other disk not in $\mathcal{I}$. The total length of the solution would then become $|\mathcal{T}_{\mathcal{I}}| + O(1) \cdot |\mathcal{I}| \cdot R$. Specifically, Dumitrescu and T\'{o}th~\cite{dumitrescu2016traveling} consider short curves around each disk in $\mathcal{I}$ that are guaranteed to cross any of the disks to its right that intersect it. Because the maximal set was chosen from left to right, that covers all the disks that could possibly intersect it. We refer the reader to ~\cite{dumitrescu2016traveling} for the detailed construction. The authors show that the length of the resulting tour $T$ is within $O(1) \cdot |\mathcal{I}| \cdot R$ of $|T_\mathcal{I}|$:
\begin{equation}\label{bound:3}
 |T| \leq |T_\mathcal{I}| + (A\cdot k + B)\cdot R,
\end{equation}
where $A = 2\cdot (\frac{\pi}{6} + \sqrt{3}-1)$ and $B = 4 - \sqrt{3}$.

Combining \ref{bound:1} and \ref{bound:3}, we upper bound the length of the solution $|T|$ in terms of $ |TSP^*_\mathcal{I}|$ as such:
\begin{align*}
&&|T| & \leq  |T_\mathcal{I}| + (A\cdot k + B)\cdot R&&\\
&&    & \leq  a \cdot |TSP^*_\mathcal{I}| + (A\cdot k + B)\cdot R &&
\end{align*}

In order to complete the analysis, we would need to bound $ |TSP^*_\mathcal{I}|$ in terms of $|TSPN^*|$ and we do that through $|TSPN^*_{\mathcal{I}}|$. The analysis from  Dumitrescu and T\'{o}th~\cite{dumitrescu2016traveling} uses the bounds from Dumitrescu and Mitchell~\cite{dumitrescu2003approximation} for the case of disjoint disks. Specifically, they apply Lemma~\ref{lem:TSPNtour} to get that:
\begin{equation*}
 kR \leq \frac{4}{\pi} \cdot |TSPN^*_{\mathcal{I}}| + 4 R.
\end{equation*}
This, together with the bound $|TSP^*_{\mathcal{I}}| \leq |TSPN^*_{\mathcal{I}}| + 2Rk$ and (\ref{bound:2}) yields:

 \begin{align*}
 &&|T| & \leq  a\cdot |TSP^*_\mathcal{I}| + (A k + B)\cdot R&&\\
 &&    & \leq  a \cdot (|TSPN^*_{\mathcal{I}}| + 2Rk) + (A k + B)\cdot R&&\\
 &&    &\leq  a \cdot |TSPN^*_{\mathcal{I}}| + (2 a + A) \cdot kR + BR&&\\
 &&    &\leq a \cdot |TSPN^*_{\mathcal{I}}| + (2 a+ A) \cdot \Big( \frac{4}{\pi} |TSPN^*_{\mathcal{I}}| + 4 R\Big)  + BR &&\\
 &&    &\leq \Big(a + (2 a + A)\frac{4}{\pi}\Big) \cdot |TSPN^*_{\mathcal{I}}| + (8a + 4 A + B)R&&\\
 &&   &\leq \Big((1+\frac{8}{\pi})a+\frac{4A}{\pi}\Big) \cdot |TSPN^*_{\mathcal{I}}| + (8a + 4 A + B)R&&\\
 &&   &\leq \Big((1+\frac{8}{\pi})a+\frac{4A}{\pi}\Big) \cdot |TSPN^*| + (8a + 4 A + B)R&&
 \end{align*}

Plugging in the values for $A$ and $B$ gives an overall approximation term of:
\begin{equation*}
 (1+\frac{8}{\pi})a+\frac{4A}{\pi} \leq \Big( \frac{7}{3} + \frac{8\sqrt{3}}{\pi}\Big) \cdot (1+\epsilon) \leq 6.75\cdot(1+\epsilon).
\end{equation*}

Our framework changes the last stage in which we compare $|TSP^*_{\mathcal{I}}|$ with $|TSPN^*_{\mathcal{I}}|$. We do a similar analysis as in the disjoint case, except for the tour on $\mathcal{I}$. We get that \textbf{Case 1} would therefore correspond to getting that:
\begin{equation*}
 |TSP^*_\mathcal{I}| \leq |TSPN^*_\mathcal{I}| + X \cdot R \cdot k,
\end{equation*}
where $ X = \displaystyle{ 2 - \frac{1 -\cos \beta }{\alpha}  }$ (instead of $2R$).
We can then replace it in the analysis and get:
 \begin{align*}
 && |T| & \leq   a \cdot (|TSPN^*_{\mathcal{I}}| + XRk) + (A k + B)\cdot R&&\\
 &&  &\leq \Big(a + (X a + A)\frac{4}{\pi}\Big) \cdot |TSPN^*_{\mathcal{I}}| + (8a + 4 A + B)R&&\\
 &&  & \leq \Big((1+\frac{4X}{\pi})a+\frac{4A}{\pi}\Big) \cdot |TSPN^*| + (4Xa + 4 A + B)R  &&
 \end{align*}

In \textbf{Case $2$}, we have that the overall detour is $2Rk$, but there is a different lower bound on $|TSPN^*_\mathcal{I}|$:
\begin{equation*}
|TSPN^*_\mathcal{i}| \geq Y \cdot Rk,
\end{equation*}
where  $Y = \frac{\alpha-1}{\alpha} \cdot \Big (1/(2\sin{(2\beta)}) -1 \Big)$. Using the fact that $Rk \leq 1/Y \cdot |TSPN^*_\mathcal{I}|$, the analysis then becomes:
\begin{align*}
 && |T|   &\leq  \alpha \cdot |TSPN^*_{\mathcal{I}}| + (2 \alpha + A) \cdot kR + BR&&\\
 &&       &\leq  \alpha \cdot |TSPN^*_{\mathcal{I}}| + \frac{2 \alpha + A}{Y} \cdot |TSPN^*_{\mathcal{I}}| + BR&&\\
 &&       &\leq \Big( \alpha + \frac{2 \alpha + A}{Y} \Big) \cdot |TSPN^*_{\mathcal{I}}|  + BR.&&
\end{align*}
If we set $\alpha $ and $\beta$ like in the previous section, we get that both of the approximation factors are upper bounded by $6.728$.

\subsection{The Straight Line Case}\label{app:line}

Here we focus on the second possibility in Theorem ~\ref{thm:main} in which the optimal TSPN is supported by a straight line that stabs all the disks. We show that in this case, we can return in polynomial time a solution that is within an additive factor of $4R$ from the optimal $TSPN^*$. We note that when the TSPN might not be a line but the disks themselves admit a line transversal, a $\sqrt{2}$-approximation follows from the work of Dumitrescu and Mitchell~\cite{dumitrescu2003approximation}. We explain the result for completeness.

We start by identifying the centers that are the farthest apart and considering the direction orthogonal to the line going through them. This direction induces parallel segments of length $2R$ in each of the disks (that each go through the centers). It is easy to check that any line transversal through the disks is a line transversal through the segments except for the first and last disk in the associated geometric permutation (for those two disks, the TSPN will stop at the boundary of the disk and never cross the entire circle). Conversely, any line transversal through the segments will automatically also stab the disks. Now compute a shortest line segment that stabs all of these segments in time $O(n\log n)$ using the algorithm of Bhattacharya et al.~\cite{bhattacharya1992computing}. We note that this is optimal up to an additive factor of $4R$ that comes from the fact that the optimal $TSPN^*$ might have to travel $4R$ to hit the first and the last two segments in the geometric permutation.

In general, when we know that the disks admit a line transversal, we can output a solution that is a $\sqrt{2}$-approximation~\cite{dumitrescu2003approximation}. This follows indirectly from an algorithm used for connected regions of the same diameter, when there is a line that stabs all of the diameters. Given the parallel segments of length $2R$ that we constructed earlier, we know that they can also be stabbed by a line. Now consider the smallest perimeter axis-aligned rectangle that intersects all of the segments, of width $w$ and height $h$. This will be the solution that we return. Arkin and Hassin~\cite{arkin1994approximation} argued that any tour which touches all four sides of the rectangle must have length at least $2\sqrt{h^2+w^2}$. Since $h+w \leq \sqrt{2}\cdot \sqrt{h^2+w^2}$, we get that the rectangle is a $\sqrt{2}$-approximation.

\section{Introduction}
We study the Traveling Salesman Problem with Neighborhoods (TSPN) when each neighborhood is a disk of fixed radius $R$. The problem is a generalization of the classical Euclidean Traveling Salesman Problem (TSP), when each point to be visited is replaced with a region (interchangeably, a neighborhood) and the objective is to compute a tour of minimum length that visits at least one point from each of these regions. While it is known that Euclidean TSP admits a Polynomial Time Approximation Scheme (PTAS) due to the celebrated results of Arora~\cite{arora1998polynomial} and Mitchell~\cite{mitchell1999guillotine}, Euclidean TSPN has been shown in fact to be APX-hard~\cite{safra2006complexity,de2005tsp} even for line segments of comparable length~\cite{elbassioni2009approximation}. The geometric version of TSPN was first studied by Arkin and Hassin~\cite{arkin1994approximation} who gave constant factor approximations for a variety of cases. Since then, there has been a wide ranging study of TSPN for different types of regions. In the case of connected regions, there is a series of $O(\log n)$ approximations~\cite{mata1995approximation, elbassioni2009approximation,gudmundsson1999fast}. Better approximations are known for cases that consider various restrictions on the regions such as comparable sizes (i.e. diameter), fatness (ratio between the smallest circumscribing radius and largest inscribing radius, or how well can a disk approximate the region) and  pairwise disjointness or limited intersection~\cite{dumitrescu2003approximation,de2005tsp,elbassioni2009approximation,mitchell2010constant,chan2011qptas,dumitrescu2016traveling, bodlaender2009minimum,mitchell2007ptas,chan2018reducing}. We refer the reader to ~\cite{handbook} for a comprehensive survey of the results.

We study the disk version which models the situation in which each customer is willing to travel a distance $R$ to meet the salesperson. This is considered an important special case of the general TSPN~\cite{dumitrescu2016traveling} and is especially relevant since it has found applicability in other areas such as path planning algorithms for coverage with a circular field of view~\cite{arkin2000approximation,galceran2013survey} and most recently for data collection in wireless sensor networks~\cite{yuan2007optimal, ma2013tour, citovsky2015exact}. Various heuristics~\cite{yuan2007optimal, carrabs2017novel,chang2015artificial,liu2013path}  and variations ~\cite{bhadauria2011robotic, ma2013tour,tekdas2012efficient} have been considered, all of which have as their basis the TSPN on uniform disks problem.

Dumitrescu and Mitchell~\cite{dumitrescu2003approximation} were the first to specifically address the case of uniform disks in 2001. They showed a PTAS for disjoint unit disks and simpler constant factor approximations for the disjoint and overlapping cases. The specific factor of $3.547$ for disjoint disks is relative to using a routine for TSP on points (i.e. the actual constant depends on the subroutine used). Later,  Dumitrescu and T\'{o}th~\cite{dumitrescu2016traveling} improved the constant factor in~\cite{dumitrescu2003approximation} for the overlapping case and extended it to unit balls in $\mathbb{R}^d$, giving a $O(7.73^d)$-approximation. When the balls are disjoint, Elbassioni et al.~\cite{elbassioni2009approximation} showed a $O(2^d/\sqrt{d})$-approximation. Most recently, Dumitrescu and T\'{o}th~\cite{dumitrescu2017constant} gave a general constant factor for (potentially overlapping) disks of arbitrary radii.  As noted by the authors, while the complexity of the disk case is well understood generally, the question of obtaining practical and better constant factor approximations remains of high  interest \cite{dumitrescu2016traveling}.

In this paper, we aim for an improved constant factor algorithm for the case of uniform radius disks and note that the algorithm proposed by Dumitrescu and Mitchell~\cite{dumitrescu2003approximation} for the disjoint case outputs an approximate TSP tour on the \textit{centers} of each disk.  The core of their analysis is a bound that compares the length of the optimal TSP on the centers of each disk($|TSP^*|$) with the length of the optimal TSPN on the disks ($|TSPN^*|$) and says that
\begin{equation}\label{bound}
|TSP^*|\leq |TSPN^*|+2Rn,
\end{equation}
where $n\geq 2$ is the number of disks in the instance. In addition, the authors use a packing argument to lower bound the length of the optimal TSPN tour in terms of $R$ and $n$ and get that $ \frac{\pi}{4}Rn - \pi R  \leq |TSPN^*| $. Overall, this gives a $3.547$- approximation and in addition, the authors show that the algorithm cannot give better than a factor $2$ approximation. While other methods for choosing representative points can be employed~\cite{dumitrescu2017constant,arkin1994approximation,dumitrescu2003approximation}, this approach is appealing both in its elegance and because it does not depend on $R$. Moreover, other existing constant factors approximations for TSPN often hide large constants~\cite{de2005tsp, elbassioni2009approximation, dumitrescu2017constant} that are incurred as a consequence of using general bounds on the length of the optimal tour that do not directly exploit the structure of the regions or of the optimal TSPN tour (bounding rectangle argument and Combination Lemma in ~\cite{arkin1994approximation}). In order to improve on them, the challenge then becomes to develop bounds that exploit the difference in behavior between  a TSP tour (on points) and the TSPN tour on the regions and furthermore, avoid using general purpose techniques that add on to the overall approximation factor.

In this context, one way to improve the approximation factor for disjoint disks is to better understand the relationship between the optimal TSP tour on the centers and the optimal TSPN on the disks. Specifically, is the $2Rn$ term in (\ref{bound}) tight or can it be improved by using specific structural properties of the optimal TSPN? A similar question was asked in 2011 by H\"{a}me, Hyyti\"{a} and Hakula~\cite{hame-eurocg-2011} for the case when $R$ is very small (and hence, $TSP^*$ and $TSPN^*$ respect the same order and the disks are pairwise disjoint). They conjectured that the true detour term should be $\sqrt{3}nR$ and constructed arbitrarily large instances of disjoint disks that converge to this case. We refer to this as the \textbf{H\"{a}me, Hyyti\"{a} and Hakula conjecture.} Subsequent experiments by M\"{u}ller~\cite{muller2011finding}, however, suggest that this might be true only for tours up to five disks and higher otherwise. No further progress has been made towards the conjecture since then.

\paragraph*{Contributions.}
We make the first progress on the conjecture and develop a twofold method that either improves the bound in (\ref{bound}) or
shows that the TSP on the centers is a good approximation for the TSPN on the disks. Formally, we get that

\begin{theorem}\label{thm:mama}
	For any $n \geq 4$ disjoint disks of radius $R$ at least one of the following is true:
	\begin{itemize}
		\item $TSPN^*$ is supported by a straight line,
		\item $|TSP^*| \leq |TSPN^*| + 1.999Rn$,
		\item $|TSP^*| \leq 2\cdot |TSPN^*|$.
	\end{itemize}
	Our framework also gives an overall $3.53$-approximation for the case of uniform disjoint disks and a $6.728$-approximation for the overlapping case.
\end{theorem}

While the improvement in the overall approximation factor is small, our framework strives to provide new insight into the problem that can be explored further. Specifically, the $2$-approximation (optimal with respect to the method of computing a TSP on the centers~\cite{dumitrescu2003approximation}) comes from the case in which the TSPN tour takes a lot of sharp turns. Furthermore, it is based on a lower bound that does not rely on packing arguments. To the best of our knowledge, this is the first such bound specifically for TSPN out of all arguments for general fat regions~\cite{mitchell2010constant}. As such, it might be of independent interest and it could, for example, lead to improved approximation factors for balls in $\mathbb{R}^d$ that do not depend exponentially on the dimension. Moreover, the fatness of the disks is used in showing that short sharp turns lead to a disk being visited multiple times and can conceivably be used to show similar properties for other fat regions.

We start by fixing an order $\sigma$ and comparing the TSPN tour that visits the disks in that order to the TSP tour that visits their respective centers in the same order. The $2Rn$ term in (\ref{bound}) comes from considering the points at which the TSPN touches the boundary of each disk and charging each such vertex with a $2R$ detour for going to its respective center and coming back. In this view, the $2Rn$ term cannot be improved since the charge on each vertex will always be $2R$. Instead, we reinterpret the bound as charging the \textit{edges} of the TSPN tour instead of its vertices and notice that the charge for each edge can now be anywhere between $-2R$ and $2R$, depending on how close the tour is to (locally) visiting pairs of disks optimally. In this context, we define a ``bad'' edge to be one that incurs a large charge (i.e. $>(2-\epsilon)R$ for some $\epsilon>0$). We show that such bad edges lead to the TSPN tour exhibiting sharp turns (i.e. with small interior angle). When the edges of the sharp turn are long, we use that to derive a better lower bound on the overall TSPN tour length. On the other hand, when one of them is short, we show that the tour must then visit a disk twice (i.e. visit it once, then touch another disk and return back to it). The crux of the argument is in understanding how these short sharp turns that visit a disk multiple times influence the global detour term.

When a tour visits a disk more than once, two scenarios follow naturally from the classical TSP case of just visiting points: either the order $\sigma$ is not optimal or the tour must follow a straight line. Surprisingly, we show that a third alternative scenario is also possible, whose local structure we call a $\beta$-triad. The main technical contribution of the paper is in describing structural properties of such $\beta$-triads and showing that they actually have a low \textit{average} detour. Specifically, we construct an additional order $\sigma'$ and use an averaging argument to show that $\beta$-triads have low detour when compared to the TSP tours that visit the centers in the order $\sigma$ and $\sigma'$. This then allows us to conclude that they have a low detour with respect to the optimal TSP on the centers.

Along the way, we also show that the H\"{a}me, Hyyti\"{a} and Hakula conjecture is true for $n=3$ and use it to bound the average detour of $\beta$-triads.  We include a discussion of the method used to derive it, involving Fermat-Weber points, which might be useful for the case of $n\geq 4$. We also discuss how our approach can be used within the framework of Dumitrescu and T\'{o}th~\cite{dumitrescu2016traveling} to yield improved approximation factors for the overlapping disks case.

\paragraph*{Preliminaries.}
We consider $n \geq 3$ disjoint disks of radius $R$ in the Euclidean plane.  We denote an optimal TSP tour on the centers of the disks as $TSP^*$. Similarly, $TSPN^*$ will denote an optimal TSPN tour on the disks. Our results will be with respect to a fixed TSPN tour (which we call simply $TSPN$) described by a sequence of ordered points $P_1, P_2, \ldots, P_n$  on the boundary of the disks such that the tour is a polygonal cycle with edges $(P_i, P_{i+1})$. Furthermore, we have that for each of the input disks, there exists some $i \in [1,n] $ such that point $P_i$ is on the boundary of the disk.

Notice that the points $P_i$ induce a natural order $\sigma$ on the disks with centers $O_1, O_2, \ldots O_n$, i.e. $\sigma$ corresponds to the identity permutation on $P_1, \ldots, P_n$. For the majority of our theorems, we will assume that $TSP$ always refers to a tour \textit{on the centers} and in the order $\sigma$ on the disks. When we need to make a difference, we will further use $TSP(\sigma')$ to be the tour which visits the centers in the order given by the permutation $\sigma'$. Given two such permutations $\sigma$ and $\sigma'$, we say that $\sigma \cap \sigma'$ refers to the maximal set of points on which $\sigma$ and $\sigma'$ agree. In this context, $TSPN(\sigma \cap \sigma')$ refers to the collection of paths we get from visiting the points $P_i$ according to $\sigma \cap \sigma'$. Similarly, $TSP(\sigma \cap \sigma')$ corresponds to the collection of paths that we get from visiting the points $O_i$ according to $\sigma \cap \sigma'$.

Finally, we denote the length of a tour $\mathcal{T}$ as $|\mathcal{T}|$. When $\mathcal{T}$ is a collection of paths, we have that $|\mathcal{T}|$ represents the total length of each of the paths. When $A$ and $B$ are points, we have that $|AB|$ denotes the length of the segment $AB$.  We therefore have that $|TSPN| = \sum_{i=1}^{n} |P_iP_{i+1}|$ and $|TSP| = \sum_{i=1}^{n} |O_iO_{i+1}|$, where $P_{n+1}=P_1$ and $O_{n+1}=O_1$.

\subsection{Consecutive Bad Edges}
\begin{figure}
\centering
\renewcommand{\baselinestretch}{1}
	\small\normalsize
  \includegraphics[height = 1.75in]{fig0}
 \caption[Description of a bad edge.]{Bad edges are guaranteed to have both endpoints in the blue arcs: if $|P_1P_2| \leq |AC|$, then $P_1P_2$ is a bad edge. In contrast, the dashed edges are guaranteed to be good edges.}\label{fig:beta}
\end{figure}
The idea behind defining bad edges in terms of $f(O_1O_2,\beta)-R$ is that it allows us to restrict the location of $P_1$ and $P_2$ on the boundary of their respective disks as seen in Figure ~\ref{fig:beta}. Specifically, there are exactly two points $A$ and $B$ on the boundary of the first disk with the property that the shortest distance from $A$ or $B$ to the boundary of the second disk is exactly  $f(O_1O_2,\beta)-R$. Not coincidentally, they form an angle of $\beta$ with $O_1O_2: \angle AO_1O_2 = \angle BO_1O_2= \beta$. In general, $P_1$ (and in a similar fashion $P_2$) is guaranteed to lie in the short arc between $A$ and $B$ whenever $P_1P_2$ is upper bounded by  $f(O_1O_2,\beta)-R$:

\begin{lemma} \label{lemma:angle} If $P_1P_2$ is bad, then the angles $\angle O_1O_2P_2$ and $\angle O_2O_1P_1$ are $\leq \beta$.
\end{lemma}
\begin{proof}
 Let $\gamma = \angle O_1O_2P_2 \leq \pi$ and notice that $O_1P_2= f(O_1O_2,\gamma)$. Consider the point $Q$ where $O_1P_2$ intersects the first disk and note that the shortest distance from $P_2$ to the first disk is exactly $P_2Q = f(O_1O_2,\gamma)-R$. We therefore get that $P_1P_2 \geq P_2Q$.
 Now notice that, if $\gamma > \beta$, then $f(O_1O_2,\gamma) > f(O_1O_2, \beta)$ and so $P_2Q > f(O_1O_2,\beta)-R$, which would lead to a contradiction. The same argument can be applied for $P_2$ and we get our conclusion.
\end{proof}

When a second bad edge $P_2P_3$ is considered, we can conclude that the angle $O_1O_2O_3$ has to be at most $2\beta$ and hence the TSP on the centers must make a sharp turn after it visits $O_2$. Specifically, let $O_3$ be the center of the disk visited next at $P_3$ and assume that the edge $P_2P_3$ is also bad. Notice that the angle $\angle O_1O_2O_3$ formed by the TSP is either $\angle O_1O_2P_2 + \angle P_2O_2O_3$ or $|\angle O_1O_2P_2 - \angle P_2O_2O_3|$. Regardless, we have that $\angle O_1O_2O_3 \leq \angle O_1O_2P_2 + \angle P_2O_2O_3$ and get the following corollary:

\begin{cor}\label{cor:angle}
If both $P_1P_2$ and $P_2P_3$ are bad edges, then the angle $\angle O_1O_2O_3$ is $\leq 2\beta$.
\end{cor}

If that happens and the disks are close to each other, we have that one of the edges of the TSP must actually intersect a disk twice. Specifically, if $|O_1O_2| \leq R/\sin(2\beta)$, then the support line for $O_2O_3$ must pass through the disk centered at $O_1$. In general, it is not true that if $O_2O_3$ intersects the first disk, we immediately get that the associated TSPN edge $P_2P_3$ must also intersect it. In our case, however, we have that the slope of $P_2P_3$ is very close to the one of $O_2O_3$ due to the fact that it is a bad edge. We use this information to show that if $O_2O_3$ does not intersect the first disk, then $P_2P_3$ cannot be a bad edge.

\begin{theorem}\label{thm:intersect}
  If $P_1P_2$ and $P_2P_3$ are bad edges and $O_1O_2 \leq R/\sin(2\beta)$, then the segment $P_2P_3$ intersects the disk centered at $O_1$.
\end{theorem}

\begin{proof}
We consider the case in which $\angle O_1O_2O_3 = \angle O_1O_2P_2 + \angle P_2O_2O_3$ and note that all the other cases are similar. We denote the two lines originating at $O_2$ that are tangent to the first circle as $\ell_1$ and $\ell_2$ such that the line $O_2O_3$ is in between $\ell_1$ and $O_2O_1$. Note that this is possible because the angle that $\ell_1$ forms with $O_2O_1$ is at least $2\beta$ (since $O_1O_2 \leq R/\sin(2\beta)$) but the angle that $O_2O_3$ forms with $O_2O_1$ is at most $2\beta$ (Corollary~\ref{cor:angle}).

Our strategy will be to first show that the segment $P_2P_3$ is contained in the wedge defined by $\ell_1$ and $\ell_2$ (Figure ~\ref{fig:intersect}). Notice that, since the wedge defines a convex space, it is enough to show that $P_2$ and $P_3$ are contained in it.

We first show that the point $P_2$ has to be in the wedge.  Let $S_1$ and $S_2$ be the points in which the segment $O_1O_2$ intersects the first and second disk. Similarly, let $T_2$ and $T_3$ be the points in which $O_2O_3$ intersects the second and third disk. We then have that $P_2$ is between $T_2$ and $S_2$.

Now we only need to show that $P_3$ is in between $\ell_1$ and $\ell_2$. We will do that by arguing that any choice of $P_3$ outside of the wedge will contradict the fact that $P_2P_3$ is a bad edge.

\begin{figure}[h]
\centering
\renewcommand{\baselinestretch}{1}
	\small\normalsize
  \includegraphics[height = 2in]{fig7}
 \caption[Argument for the proof of Lemma 4.2]{When $O_2O_3$ crosses the disk centered at $O_1$, we must also have that the segment $P_2P_3$ also crosses it. We show this by arguing that $P_2P_3$ is contained between the two lines $\ell_1$ and $\ell_3$ and that $P_2$ and $P_3$ are on separate sides of the first disk.}\label{fig:intersect}
\end{figure}

 Let $\alpha_1 = \angle O_1O_2P_2 $ and $\alpha_2 =  \angle P_2O_2O_3$, and so $\alpha_1, \alpha_2 \leq \beta$ (Lemma~\ref{lemma:angle}). First notice that if neither $\ell_1$ nor $\ell_2$ intersect the third disk, then we are done because we have that the entire boundary is contained in the convex space (since $O_3$ is already in between $\ell_1$ and $\ell_2$). Assume then that $\ell_1$ intersects the third disk at a point $P$ above the line $O_2O_3$, since $\angle O_3O_2O_1 \leq 2\beta \leq \angle PO_2O_1$. Moreover, since $\angle PO_2O_1 \geq 2\beta$, we have that $\angle PO_2P_2 \geq 2\beta - \alpha_1 \geq \beta$ and so $PP_2 \geq f(PO_2,\beta)$ (because $|P_2O_2| = R$). Since $|PO_2|\geq |T_3O_2|$ and $|T_3O_2| \geq R$, this implies that $|PP_2| \geq f(T_3O_2,\beta)$. Using the fact that $f(x,\beta) \geq x-R\cos \beta$ for any $x$ and $\beta \neq 0$, one can verify that:
 \begin{align*}
 f(T_3O_2,\beta) &= f(O_2O_3 -R, \beta) \\
                  &= \sqrt{(|O_2O_3|-R)^2 + R^2 - 2R(|O_2O_3|-R)\cos \beta}\\
                  &> \sqrt{|O_2O_3|^2 + R^2 - 2R|O_2O_3|\cos \beta}-R\\
                  &> f(O_2O_3,\beta)-R.
 \end{align*}
This means that $P$ cannot be a possible position for $P_3$ because then $P_2P_3$ would be too big. Moreover, any point $Q$ "above" $P$ (i.e. such that $\angle O_3O_2Q > \angle O_3O_2P$) would also not work as a possible position for the same reason. In other words, $P_3$ has to be underneath the line $PO_2 = \ell_1$.

In order to prove that $P_3$ is also above the line $\ell_2$, we will consider an additional line $\ell_3$ originating at $O_2$ that makes an angle of $\beta$ with $O_2P_2$ and is underneath it. This new line makes an angle of $\beta+\alpha_2$ with $O_2O_3$ and since  $\ell_2$ makes an angle of $\geq 2\beta+\alpha_1+ \alpha_2$ with $O_2O_3$, we get that $\ell_3$ is in between $O_2O_3$ and $\ell_2$. In other words, if we show that $P_3$ is above $\ell_3$, then we also get that $P_3$ is above $\ell_2$. If $\ell_3$ does not intersect the third disk, then we are done as before, so assume that it intersects it at a point $Q$ on the boundary. Similarly as before, we have that $P_2Q = f(O_2Q,\beta) \geq f(O_2T_3,\beta) > f(O_2O_3,\beta)-R$. This in turn implies that $P_3$ has to be above $Q$, otherwise $P_2P_3$ would be too big. Therefore $P_3$ must be above the line $\ell_3$.

At this point, we have that the segment $P_2P_3$ is contained in the wedge defined by $\ell_1$ and $\ell_2$. We know that the first disk is tangent on both sides to $\ell_1$ and $\ell_2$ but this does not directly imply that $P_2P_3$ must actually intersect it. In order to have that, we must also ensure that $P_2$ and $P_3$ lie on different sides of the first disk. We argue this by showing that $O_3$ itself must be on the other side of the first disk as $O_2$. Since the disks do not intersect, this implies that $P_3$ is on a different side from $P_2$. In order to show this, notice that we can assume, without loss of generality, that $O_1O_2 \leq O_2O_3$. Let $T$ be the point on $O_2O_3$ such that $O_1T \perp O_2O_3$. Since $\angle O_1O_2O_3 \leq 2\beta$ and  $O_1O_2 \leq R/\sin(2\beta)$, this means that $T$ is contained in the first disk. Suppose that $O_3$ is on the segment $O_2T$ (effectively in between $O_1$ and $O_2$). Then $O_2O3 < O_2T$ but, since $O_2T = O_1O_2 \cos(\angle O_1O_2T) \leq O_1O_2$, this would lead to a contradiction. We therefore get that $O_2$ and $O_3$ are on different sides of the first disk and that the same is true for $P_2$ and $P_3$. This shows that the segment $P_2P_3$ must intersect the first disk.
\end{proof}

\subsection{Introducing $\beta$-triads}

The fact that the disk centered at $O_1$ is crossed by both $P_1P_2$ and $P_2P_3$ suggests that the TSPN might not be optimal because it could be shortcut. Our structural theorem identifies when that is the case and isolates the remainder as having a specialized local structure which we call a $\beta$-triad. Formally, we say that a specific TSPN subpath  $P_n - P_1 - P_2 - P_3$ is a \textbf{$\beta$-triad} if it satisfies all of the following properties (Figure~\ref{fig:triadpic}):
\begin{itemize}
\item $P_1P_2$ and $P_2P_3$ are bad edges and $O_1O_2 \leq R/\sin(2\beta)$,
\item $P_1, P_2, P_3$ are not collinear but $P_n,P_1,P_2$ are collinear with $P_1$ between $P_n$ and $P_2$.
\end{itemize}

\begin{figure}[h]
\centering
\renewcommand{\baselinestretch}{1}
	\small\normalsize
  \includegraphics[height = 2in]{fig9}
 \caption[Definition of a $\beta$-triad.]{The path $P_n - P_1 - P_2 - P_3$ forms a $\beta$-triad.}\label{fig:triadpic}
\end{figure}

We state the structural theorem here and refer the reader to Appendix~\ref{app:main} for a complete argument. The case in which the TSPN tour follows a straight line that stabs all the disks is discussed separately in Section BLAH BLAH BLAH and is of separate interest.

\begin{theorem}\label{thm:main}
 For $n\geq 4$, if $P_1P_2$ and $P_2P_3$ are bad edges  and $O_1O_2 \leq R/\sin(2\beta)$ then at least one of the following is true:
 \begin{itemize}
 \item the TSPN tour is not optimal,
 \item the TSPN tour is supported by a straight line or
 \item the path $P_n - P_1 - P_2 - P_3$ forms a $\beta$-triad.
 \end{itemize}
\end{theorem}

\begin{proof}
 We distinguish between the case in which $P_2P_3$ intersects the first disk at $P_1$ and otherwise. In the first case, we will show that either the TSPN is not optimal or all the disks are stabbed by it. The second case is more involved and reduces to describing what the local structure of the TSPN must be such that it does not necessarily fall in the previous two cases.

\textbf{Case $1$: $P_1,P_2,P_3$ are collinear.} Then consider the point $P_n$ that connects to $P_1$. The cost that the TSPN pays for visiting the four disks is $|P_nP_1|+|P_1P_2|+|P_2P_3|$ but by triangle inequality, we know that $|P_nP_2| \leq |P_nP_1| + |P_1P_2|$, so the TSPN would visit $P_2$ directly and pass through $P_1$ on its way to $P_3$. If the inequality is strict, then this directly implies that the TSPN is not optimal. When we have equality, however, this implies that $P_n,P_1$ and $P_2$ are now also collinear and furthermore, that $P_1$ lies between $P_2$ and $P_n$. In other words, we have that on the line from $P_2$ to $P_n$, we have both $P_3$ and $P_n$ to the left of $P_1$. Now look at how point $P_4$ connects to $P_3$ and notice that the portion of TSPN for the five disks is now $|P_4P_3| + |P_3P_2|+|P_1P_2|+|P_1P_n|$ and again, we can ask the question of why wouldn't the TSPN go straight to $P_2$ instead and visit $P_3$ along the line $P_2P_3$. Specifically, we have $|P_4P_2| \leq |P_4P_3| + |P_3P_2|$ with the TSPN not being optimal whenever this inequality is strict. We therefore consider the case in which $|P_4P_2|= |P_4P_3| + |P_3P_2|$ and get that now $P_4$ has to also be collinear with the other points and furthermore, $P_3$ has to be between $P_4$ and $P_2$. Continuing this process, we get that all the TSPN points would have to be collinear and in the order $P_2,P_1,P_3,P_4,\ldots, P_{n-1}$ with $P_n$ potentially being anywhere past $P_1$. In this case, we have that the TSPN is a supported by a straight line that stabs all of the disks.

\textbf{Case $2$: $P_1,P_2,P_3$ are not collinear.} Let the line $P_1P_2$ intersect the first disk for the first time at $Q_1$. By the argument from before, we know that if $|P_nP_2| < |P_nP_1| + |P_1P_2|$, then the TSPN cannot be optimal since another tour could go from $P_n$ straight to visiting $P_2$ and then visit $P_1$ on the way to $P_3$, at a lesser cost. When $P_n,P_1$ and $P_2$ are collinear, in that order, we say that $P_n - P_1 - P_2 - P_3$ form a $\beta$-triad.

\end{proof}

\subsection{Properties of $\beta$-triads}
Theorem ~\ref{thm:main} says that if $|TSPN^*|$ is not a straight line, then the triad has a local detour of at most $3\sqrt{3}R$. Lemma ~\ref{lem:betatriad} further states that all the bad triads are also edge disjoint.  In order to prove that, we go back to the proof of Theorem~\ref{thm:main}. Note that we distinguished between the case in which $P_1,P_2$ and $P_3$ are collinear (Case $1$) and when they are not(Case $2$). The first case leads to the TSPN being a straight line, which is ruled out by our assumptions. In the second case, the optimality of $TSN^*$ implies that $P_n, P_1$ and $P_2$ are also collinear, with $P_1$ between $P_2$ and $P_n$.

\begin{lemma}\label{lem:betatriad}
All the $\beta$-triads in a given TSPN tour are edge disjoint.
\end{lemma}

\begin{proof}
Assume there is another bad triad that shares edges with $P_n - P_1 - P_2 - P_3$. We distinguish four cases, based on the type of edges they have in common.

\textbf{Case $1$: $P_{n-1} - P_n - P_1 - P_2$ is a bad triad.}
This case cannot happen since $P_n,P_1,P_2$ are collinear.

\textbf{Case $2$: $P_{n-2} - P_{n-1} - P_n - P_1$ is a bad triad.} Then, by definition, we must have that $P_nP_1$ is also a bad edge. We will show, however, than this cannot be. For this, we will use an additional lemma:

\begin{figure}[h]
\centering
\renewcommand{\baselinestretch}{1}
	\small\normalsize
  \includegraphics[height = 2in]{fig8}
 \caption[Argument for the proof of Lemma 4.2.]{A potential TSPN path is drawn in blue. Because the angle $\angle O_iO_{i+1}O_{i+2}  \leq \pi/2$, we have that $O_i$ and $O_{i+1}$ are on the same side of the hyperplane described by the red line. That, in turns, give two options for the disk centered at $O_{i+2}$ to intersect the (extended) convex hulls of the other two diskss, drawn in green. In each case, the points are visited in the wrong order.}\label{fig:convexhull}
\end{figure}

\begin{lemma}\label{lem:disjoint} If $P_iP_{i+2}$ is a straight line that passes through point $P_{i+1}$ such that $P_{i+1}$ is between $P_i$ and $P_{i+2}$, then it cannot be that both $P_iP_{i+1}$ and $P_{i+2}P_{i+1}$ are bad edges.
\end{lemma}
\begin{proof}
Assume that both $P_iP_{i+1}$ and $P_{i+2}P_{i+1}$ are bad edges. Then Corollary ~\ref{cor:angle} implies that the angle $\angle O_{i}O_{i+1}O_{i+2} \leq 2\beta$. Now consider the convex hull of the two disks centered at $O_i$ and $O_{i+1}$ (Figure ~\ref{fig:convexhull}). If the disk centered at $O_{i+2}$ intersects the convex hull, then $P_{i+2}$ must be contained in that convex hull, otherwise the line $P_i - P_{i+1} - P_{i+2}$ would not exist. But in that case, the points would be visited out of order. Specifically, $P_{i+2}$ would be between $P_i$ and $P_{i+1}$.

Now assume that the disk centered at $O_{i+2}$ does not intersect the convex hull. Since the angle $\angle O_{i}O_{i+1}O_{i+2} \leq 2\beta \leq \pi/6$, this implies that $O_{i+2}$ is in the same halfspace as $O_{i}$ with respect to the line perpendicular to $O_iO_{i+1}$ passing through $O_{i+1}$. We extend the convex hull infinitely in that halfspace by allowing the tangent lines to be infinite on that side. By the same argument as before, we know that the disk centered at $O_{i+2}$ must intersect this extended region. But then we would get again that the points are out of order: $P_{i}$ would be between $P_{i+1}$ and $P_{i+2}$.
\end{proof}

When $P_i = P_n, P_{i+1} = P_1$ and $P_{i+2} = P_2$, Lemma ~\ref{lem:disjoint} tells us that it cannot be that $P_1$ is between $P_n$ and $P_2$ and both edges $P_1P_n$ and $P_1P_2$ are bad. Therefore we are done with this case.

\textbf{Case $3$: $P_1 - P_2 - P_3 - P_4$ is a bad triad.} This case is similar to Case $1$ and cannot happen, since $P_1, P_2$ and $P_3$ cannot be collinear.
\textbf{Case $4$: $P_2 - P_3 - P_4 - P_5$ is a bad triad.} This case is similar to Case $2$ because we have that $P_2$, $P_3$ and $P_4$ are collinear with $P_3$ between $P_2$ and $P_4$ and both $P_2P_3$ and $P_3P_4$ being bad edges.

\end{proof}

\subsection{Structural Theorem}
The case of $\beta$-triads is interesting because it arises naturally as a consequence of dealing with regions instead of points. Given any optimal tour that exhibits internal angles $\leq \pi/6$, we can always add an extra disk at each sharp turn that will maintain optimality, pairwise disjointness and be intersected twice by this tour, giving rise to a $\beta$-triad. It is therefore important that we understand their behaviour.

Because of the fact that $P_1P_2$ and $P_2P_3$ are bad edges, the $\beta$-triad is likely to have a high detour with respect to $TSP(\sigma)$. Nevertheless, we show that there exists an alternate ordering $\sigma'$ such that the average detour of the three edges in the $\beta$-triad with respect to $(|TSP(\sigma)| + |TSP(\sigma')|)/2$ is $3\sqrt{3}R$. The order $\sigma'$ takes advantage of the fact that the disk centered at $O_1$ is crossed twice and inverts the order in which it is visited without changing the cost of the underlying TSPN tour. The $3\sqrt{3}R$ bound comes from proving the H\"{a}me, Hyyti\"{a} and Hakula conjecture for $n=3$ (Section ~\ref{fermat}). In order to be able to construct $\sigma'$ consistently across multiple $\beta$-triads, we also show that $\beta$-triads are isolated events and specifically that they are edge-disjoint (Lemma ~\ref{lem:betatriad}).

\begin{theorem}\label{thm:triadbound} If the TSPN in the order $\sigma$ has $k$ $\beta$-triads that together cover a set of edges of total length $L_T$, then we can construct another order $\sigma'$ that agrees with $\sigma$ on everything except the order inside the $\beta$-triads such that:
\begin{center}
 $\displaystyle{ \frac{|TSP(\sigma)| + |TSP(\sigma')|}{2} \leq |TSP(\sigma \cap \sigma')| + L_T+ 3\sqrt{3}Rk}$.
\end{center}
\end{theorem}

\begin{proof}
We discuss the case for $k=1$ and show how to modify the argument for $k>1$. Suppose that $P_n - P_1 - P_2 - P_3$ form a $\beta$-triad. We know that $TSP(\sigma)$ visits the centers of each disk in the order $O_1, O_2, O_3, \ldots O_n$. We consider an additional order $\sigma'$ such that $TSP(\sigma')$ visits the centers in the order $O_2, O_1,O_3, \ldots O_n$. Notice that both $\sigma$ ad $\sigma'$ agree on the order $O_3,O_4, \ldots, O_{n}$ and that they differ in the fact that $\sigma$ visits $O_1$ after $O_n$ and before $O_2$ and $\sigma'$ visits $O_1$ after $O_2$ and before $O_3$. Therefore,  for $T' =TSP(\sigma \cap \sigma')$, we have that  $|T'| = |O_3O_4|+\ldots+|O_{n-1}O_n|$,  $|TSP(\sigma)| = |T'| +  |O_nO_1| + |O_1O_2| + |O_2O_3|$ and $|TSP(\sigma')| = |T'| +  |O_nO_2| + |O_2O_1| + |O_1O_3|$.

On the other hand, the length of the TSPN with respect to the orders $\sigma$ and $\sigma'$ stays the same. The local cost of visiting $P_n - P_1 - P_2 - P_3$ is $L_T = |P_nP_1| + |P_1P_2| + |P_2P_3| = |P_nP_2| + |P_2P_3|$, since $P_n, P_1$ and $P_2$ are collinear and $P_1$ is between $P_n$ and $P_2$. We also know that $P_2P_3$ intersects the disk centered at $O_1$ at some point $Q_1$ that is different from $P_1$ (Theorem ~\ref{thm:intersect}). In other words, the TSPN that visits the points $P_n - P_1 - P_2 - P_3$ can be reimagined as visiting the points $P_n - P_2 - Q_1 - P_3$ and therefore respecting the order $\sigma'$. The local cost of crossing these edges is the same as before: $|P_nP_2| + |P_2Q_1| + |Q_1P_3| = |P_nP_2| + |P_2P_3| = L_T$.

We now apply Theorem~\ref{thm:triangle} (the $3\sqrt{3}R$ bound for $n=3$) on the TSP tour $O_n -O_1 - O_2$ with the TSPN tour $P_n - P_1 - P_2$ and get that:
\begin{align*}
  |O_nO_1| + |O_1O_2| + |O_nO_2| &\leq |P_nP_1| + |P_1P_2| + |P_nP_2| + 3\sqrt{3}R &&\\
                          &\leq 2|P_2P_n| + 3\sqrt{3}R.&&
\end{align*}
On the other hand, if we consider the tour $O_1 - O_2- O_3$ with the TSPN tour $P_2 - Q_1 - P_3$, we get that:
\begin{align*}
 |O_1O_2| + |O_2O_3| + |O_1O_3| &\leq |Q_1P_2| + |P_2P_3| + |P_3Q_1| +  3\sqrt{3}R && \\
                          &\leq 2|P_2P_3| + 3\sqrt{3}R.&&
\end{align*}
Combining the two inequalities and rearranging some terms gives us that:
\begin{align*}
|TSP(\sigma)|+ |TSP(\sigma')| =& 2|T'| +  |O_nO_1| + |O_1O_2| + |O_2O_3| + \\
                               &+ |O_nO_2| + |O_2O_1| + |O_1O_3| \\
                              =& 2|T'|  + |O_nO_1| + |O_1O_2| + |O_nO_2| + \\ &+|O_1O_2| + |O_1O_3|+|O_2O_3| \\
         \leq & 2|T'|  + 2|P_2P_n| + 2 |P_2P_3| + 6\sqrt{3}R.
\end{align*}
Since $L_T = |P_2P_n| + |P_2P_3|$, we get our conclusion.

When $k>1$, we construct the order $\sigma'$ by switching the order in which we visit the centers in each $\beta$-triad in the same way as before. Since all the $\beta$-triads are edge disjoint (Lemma ~\ref{lem:betatriad}), we can construct $\sigma'$ without any conflicts because any reordering that happens in one $\beta$-triad will not affect another $\beta$-triad.
\end{proof}

redpajama_set_name: RedPajamaArXiv


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Now margrethe vestager is preparing for her final act facebook blocked 115 accounts suspected of inauthentic behavior related to u. Heres how to fix that, according to an expert nancy pelosi inevitably becomes a meme with post-oval office meeting sunglasses moment al gore says president trumps climate change stance could be a perverse step forward french protesters just won concessions from macron. Use of this site constitutes acceptance of our time may receive compensation for some links to products and services on this website. Its past time we change how americans save for retirement cdc confirms first death related to salmonella outbreak linked to raw turkey she took on silicon valley giants. Prime minister theresa may has survived a leadership challenge and will remain in her position as the country navigates the complex and politically fraught brexit process i hate being the story. Their neighbors let it happen malfunctioning heater sparked reports of active shooter at north carolina high school google doodle celebrates the life and work of native american woodcarver amanda crowe star-studded malibu forced to evacuate after threat from ferocious california wildfire president trump signs order blocking asylum claims for migrants who enter illegally macron seizes historical moment to redefine his presidency in opposition to trump dutch man, 69, launches legal battle to lower his age 20 years so he can get a date on tinder why the midterm results should concern you regardless of which party you support why it would be a major military mistake to deploy troops to confront the migrant caravans the 401(k) is turning 40 years old.
The Democratic Party is the oldest voter-based political party in the world and the oldest existing political party in the United States, tracing its heritage back to the anti-Federalists and the Jeffersonian Democratic-Republican Party of the 1790s.
History And Politics Personal Statement History and Politics Personal Statement 6Many talk of the merits of History and Politics but my reasons for studying these subjects are not . The complex relation between individual, family, society and state enthralls me. The dynamic . History and Politics Personal Statement 1History has always been an interest of mine, however my first experience studying it academically . Parents who lost children in school shootings find comfort in a group no one wants to join austrias young chancellor sebastian kurz is bringing the far-right into the mainstream inside a bangladeshi journalists dangerous journey from photographer to prisoner the yellow vests expose how fragile emmanuel macrons hold on france really is your brain works against you when you argue with your significant other. But covering philippine president duterte might be her most dangerous job yet the striking similarities between why trumps and nixons personal lawyers were sent to prison the world moves on and you dont.
The capital gazette reflects on life after a newsroom shooting she reported from war zones. Googles redesigned homepage is encouraging people to head to the polls sign up to receive the top stories you need to know now on politics, health, money and more you. Use of this site constitutes acceptance of our time may receive compensation for some links to products and services on this website. . Now margrethe vestager is preparing for her final act facebook blocked 115 accounts suspected of inauthentic behavior related to u.
Prime minister theresa may has survived a leadership challenge and will remain in her position as the country navigates the complex and politically fraught brexit process i hate being the story. At least 12 people were killed, including a sheriffs deputy who charged into the scene, after a gunman opened fire in a busy bar wednesday night in thousand oaks, calif. Here are their stories he survived the las vegas shooting only to be killed in the borderline bar 1 year later guns have divided america. But covering philippine president duterte might be her most dangerous job yet the striking similarities between why trumps and nixons personal lawyers were sent to prison the world moves on and you dont. Their neighbors let it happen malfunctioning heater sparked reports of active shooter at north carolina high school google doodle celebrates the life and work of native american woodcarver amanda crowe star-studded malibu forced to evacuate after threat from ferocious california wildfire president trump signs order blocking asylum claims for migrants who enter illegally macron seizes historical moment to redefine his presidency in opposition to trump dutch man, 69, launches legal battle to lower his age 20 years so he can get a date on tinder why the midterm results should concern you regardless of which party you support why it would be a major military mistake to deploy troops to confront the migrant caravans the 401(k) is turning 40 years old. Heres how to fix that, according to an expert nancy pelosi inevitably becomes a meme with post-oval office meeting sunglasses moment al gore says president trumps climate change stance could be a perverse step forward french protesters just won concessions from macron. Heres why theyre more successful than american demonstrators why narendra modis ruling party no longer looks invincible in 2019 indian elections a promising way to help drug users is severely lacking around the world, report says the nih says it will spend up to 20 million to find alternatives to human fetal tissue always forgetting important things? Heres how to fix that, according to science sign up to receive the top stories you need to know now on politics, health, money and more you. Its past time we change how americans save for retirement cdc confirms first death related to salmonella outbreak linked to raw turkey she took on silicon valley giants. Parents who lost children in school shootings find comfort in a group no one wants to join austrias young chancellor sebastian kurz is bringing the far-right into the mainstream inside a bangladeshi journalists dangerous journey from photographer to prisoner the yellow vests expose how fragile emmanuel macrons hold on france really is your brain works against you when you argue with your significant other. Heres what happens when 245 people try to meet in the middle jewish germans had their lives destroyed by nazis during kristallnacht.
The capital gazette reflects on life after a newsroom shooting she reported from war zones. At least 12 people were killed, including a sheriffs deputy who charged into the scene, after a gunman opened fire in a busy bar wednesday night in thousand oaks, calif. Googles redesigned homepage is encouraging people to head to the polls sign up to receive the top stories you need to know now on politics, health, money and more you. Heres what happens when 245 people try to meet in the middle jewish germans had their lives destroyed by nazis during kristallnacht. Now margrethe vestager is preparing for her final act facebook blocked 115 accounts suspected of inauthentic behavior related to u.


text:
Зубні́ при́голосні () — у фонетиці група приголосних звуків, що мають однакове місце творення: вони артикулюються язиком проти верхніх і (або) нижніх зубів. Інша назва — дента́льні при́голосні.

Зубні приголосні
[n̪] (твердий н) — дзвінкий зубний носовий
[l̪] (твердий л) — дзвінкий зубний боковий апроксимант
[r̪] (твердий р) — дзвінкий зубний дрижачий
[t̪] (твердий т) — глухий зубний проривний
[d̪] (твердий д) — дзвінкий зубний проривний
[t̪͡s̪] (твердий ц) — глухий зубний африкат
[d̪͡z̪] (твердий дз) — дзвінкий зубний африкат
[s̪] (твердий с) — глухий зубний сибілянти
[z̪] (твердий з) — дзвінкий зубний сибілянти

Джерела

Посилання

Приголосні звуки

redpajama_set_name: RedPajamaWikipedia


text:
Obituary - John Benton
The ASC are sad to announce the passing of John Benton, its first Chairman. John started at Chubb Lock & Safe and later becoming responsible for their Western Hemisphere business. Later he joined the Stratford Safe Co owned by Walter Runcimen. At about the same time Runcimen acquired John Tann the oldest safe maker in the UK where John became Managing Director. He was also a Director of Killby-Tann Security – a jointly owned company with Killby & Gayford.John gave his support and experience to many organisations such as British Standards (Bsi) where he chaired the Committee on Secure Enclosures, also the American Association for Industrial Security (ASIS). He was a member of the Editorial Committee of the 'Journal of Security'. He was a Fellow of the Institute of Directors and also he was made a Freeman of the City of London for services to Financial Institutions.
Following a spell with Schroder's the Banking giant John turned to Consultancy and it was in 1991 he, together with Mike Cahalane and Mike Collier set up the ASC, an organisation which represents the interest of Independent Security Consultants. Initially with just seven members it now, some 25 years later, has over 200 members. John, the first ASC Chairman, served five years in that position. He was instrumental in many ways, helping to form the ASC, and his links with the political world lead to him introducing early patrons Lord Nicholas Bethel and Sir Rhodes Boyson.
After a fall on Christmas evening,Johnwas admitted to St. Mary's Hospital Paddington where he died in the early evening of Thursday 16th February at the age of 91, having contracted Bronchopneumonia. Our thoughts are with his wife, children and grandchildren.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Java Android Developing "Failure in SSL library, usually a protocol" I'm programming an Application just for testing with Android Studio. I want to Request a Server with POST params which are in the string: (strings[0]).
But I always get this Exception:

javax.net.ssl.SSLProtocolException: SSL handshake aborted: ssl=0x6f427ff108: Failure in SSL library, usually a protocol error
  error:1000042e:SSL routines:OPENSSL_internal:TLSV1_ALERT_PROTOCOL_VERSION (external/boringssl/src/ssl/tls_record.cc:587 0x6fa8195d88:0x00000001)

I tried also with httpok.
https://square.github.io
The Server works with TLSV1.2
[...]
                byte[] postData = strings[0].getBytes(StandardCharsets.UTF_8);
                int postDataLength = postData.length;

                URL url = new URL("https://example.com");

                HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();

                //
                conn.setDoOutput(true);
                conn.setInstanceFollowRedirects(false);
                conn.setRequestMethod("POST");
                conn.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
                conn.setRequestProperty("charset", "utf-8");
                conn.setRequestProperty("Content-Length", Integer.toString(postDataLength));
                conn.setUseCaches(false);

                //
                // TLSv1 | TLSv1.1 | TLSv1.2
                SSLContext sc = SSLContext.getInstance("TLSv1.2");
                sc.init(null, null, new java.security.SecureRandom());
                sc.getProtocol();
                sc.getSupportedSSLParameters();
                sc.getDefaultSSLParameters();
                SSLEngine engine = sc.createSSLEngine();

                conn.setSSLSocketFactory(sc.getSocketFactory());

                conn.connect();
                try (final InputStream in = conn.getInputStream()) {
                    //…
                }
[...]

I don't get any further.

A: This is a problem in establishing the connection between the device and the mobile. Mainly happened because of used TLS version that the server support and use and the device connected client software isn't supported. For example I faced the same issue with my application I set my Nginx server configuration to use only TLS v1.3 while the okhttpclient on old android OS versions like android 8 in my case 8 supports TLS v1.2 so you have two solutions here:

*

*To change the server TLS version (which I don't recommend for you )

*Try to make okhttp work with TLSv1.3 take a look here How to enable TLSv1.3 for OkHttp 3.12.x on Android 8/9?

A: [...]
                byte[] postData = strings[0].getBytes(StandardCharsets.UTF_8);
                int postDataLength = postData.length;

                URL url = new URL("https://example.com");
  // The url has to be set to an international variable

                new Thread(new Runnable() {
                     public void run() {

                        // The url has to be fetched from an international variable
                        HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();

                //
                        conn.setDoOutput(true);
                        conn.setInstanceFollowRedirects(false);
                        conn.setRequestMethod("POST");
                        conn.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
                        conn.setRequestProperty("charset", "utf-8");
                        conn.setRequestProperty("Content-Length", Integer.toString(postDataLength));
                        conn.setUseCaches(false);

                        //
                        // TLSv1 | TLSv1.1 | TLSv1.2
                        SSLContext sc = SSLContext.getInstance("TLSv1.2");
                        sc.init(null, null, new java.security.SecureRandom());
                        sc.getProtocol();
                        sc.getSupportedSSLParameters();
                        sc.getDefaultSSLParameters();
                        SSLEngine engine = sc.createSSLEngine();

                        conn.setSSLSocketFactory(sc.getSocketFactory());

                        conn.connect();
                      });
                    }
                }).start();

                try (final InputStream in = conn.getInputStream()) {
                    //…
                }
[...]

https://developer.android.com/guide/components/processes-and-threads

redpajama_set_name: RedPajamaStackExchange


Imgessel have qualified specialists, designing your API 650, 620 or AWWA tank.
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text:
HHS Public Health Makes Statement on COVID-19 Booster Shots
Pharmaceutical Technology Editors
HHS Public Health and a team of medical experts recommend that COVID-19 vaccine booster shots will soon be necessary.
On August 18, 2021, the US Department of Health and Human Services (HHS) Public Health and a team of medical experts produced a statement on COVID-19 booster shots, saying that the COVID-19 vaccines authorized in the United States are effective, but that a booster shot will be needed to maximize vaccine-induced protection and prolong its durability. Starting in September 2021, the officials hope to begin administering booster shots to individuals who received their second dose eight months ago—pending an FDA evaluation and Centers for Disease Control and Prevention's (CDC's) issuing booster dose recommendations.
The HHS Public Health and the team of medical experts said in the statement that the COVID-19 vaccines are effective in reducing the risk of severe disease, hospitalization, and death—including against the Delta variant. However, as HHS Public Health and the team of medical experts study the scientific data surrounding the pandemic, they have found that protection against the SARS-CoV-2 infection decreases over time following the initial doses of vaccination. They also note that many vaccines (outside of the COVID-19 vaccine) have a reduction in protection over time, and additional vaccine doses may be needed to provide long-lasting protection.
"Based on our latest assessment, the current protection against severe disease, hospitalization, and death could diminish in the months ahead, especially among those who are at higher risk or were vaccinated during the earlier phases of the vaccination rollout," said HHS Public Health and a team of medical experts in the statement. "For that reason, we conclude that a booster shot will be needed to maximize vaccine-induced protection and prolong its durability."
According to this statement, HHS Public Health and the team of medical experts have developed a plan to offer COVID-19 booster shots this fall, starting September 20, 2021. Individuals will receive the booster shot eight months after their second dose. This plan is dependent upon:
FDA conducting an independent evaluation and determination of the safety and effectiveness of a third dose of the Pfizer and Moderna messenger RNA vaccines
CDC's Advisory Committee on Immunization Practices issuing booster dose recommendations
The statement is attributed to Rochelle Walensky, director, CDC; Dr. Janet Woodcock, acting commissioner, FDA; Vivek Murthy, US surgeon general; Dr. Francis Collins, director of the National Institutes of Health; Anthony Fauci, chief medical advisor to President Joe Biden and director of the National Institute of Allergy and Infectious Diseases; Rachel Levine, assistant secretary for Health; David Kessler, chief science officer for the COVID-19 response; and Marcella Nunez-Smith, chair of the COVID-19 Health Equity Task Force.
Source: HHS.gov
Bio/Pharma NewsTop NewsRegulatory Authority ActionsCOVID-19 Update
European Commission Approves Lumykras (Sotorasib) to Treat KRAS G12C-Mutated Advanced Non-Small Cell Lung Cancer


game or parts after warranty work is completed.
We cannot be held responsible for any damage due to reinstallation and/or use.
The exchange of any item or part must follow procedures and Guidelines, must be followed by the purchaser.
Used or Refurbished items are final and cannot be returned or refunded. E-proms and electrical parts are sold "As Is" and not returnable for exchange or refunds Game Room Toys / Vegas Style Slots cannot be certain these items will be installed correctly.
It is the costumer's sole responsibility to inspect all shipments when being delivered and inspect all shipping containers for damages prior to unpacking item.The customer must take pictures of damaged shipping containers/cartons.
Report any damages to us immediately! Report shipping related damage directly to the shipping or Freight Company and also notify us so there is no delays.
Notification of damages must be made to us within 5 days of receipt of damaged goods. We must recieve pictures of the damage.
And customer is responsible for notifying the delivery company of shipping damages. Keep all shipping materials as they will be needed to show the shipping company. It is very important that you notify us immediately of any damages so we can properly advise you. To speed the damage goods claim process you can e-mail us with pictures showing the damages found to the item for shipping container/carton. Upon receipt of the pictures we will arrange for a replacement to be shipped and we will advise you what to do with the damage item.
NOTE: We do not trade/swap machines do to you not liking it after you play it for awhile. Once you buy it it is yours. We will swap out a machine of the same value if there is a problem with it and it can't be fixed within the 30 day warranty period.


"Located Along the Platte River Bottoms, in Nebraska"
Safe Haven Refuge, est.1999, is a non-profit, volunteer organization dedicated to the education of prospective bird owners, and offering up-to-date and proper basic care to current bird owners. We serve in cooperation with other organizations, locally and across the U.S., in providing a safe avenue for all birds dealing with re-homing, handicapp or aggression issue.
Welcome to the Feather Picking Website!


text:
fronting up
Harry Maguire defends place in England team with Euro 2020 performance and suggests Manchester United criticism is 'for clicks' in defiant talkSPORT interview after Nations League defeat to Italy
Sam May
Updated: 24th September 2022, 10:53 am
Harry Maguire gave a defiant interview on talkSPORT to defend his place in the England team after the 1-0 defeat to Italy on Friday night.
The Three Lions were relegated from their UEFA Nations League group following Giacomo Raspadori's stunning 68th-minute strike.
Maguire has defended England and his place in the team
There was simply nothing to get excited about from an England perspective, with the latest goalless show leaving England 450 minutes without a goal in open play – their worst run in 22 years.
The Three Lions are one of only two nations who have failed to score a single non-penalty goal in the Nations League. The other is San Marino, who are made up of part-timers.
Despite a solid performance at the San Siro, Maguire's place in the starting XI has been scrutinised after he was dropped by Manchester United boss Erik ten Hag.
The Red Devils captain has endured a prolonged run of poor form – but Southgate looks set to stick with him for the World Cup in November.
When asked about his situation, Maguire told talkSPORT: "I am all good, ready to play, I feel fit and fresh.
"On my form, I don't know what has been said, I don't really read about it.
Most read in Football
The 29-year-old played the full 90 minutes in the defeat on Friday night
"I came off the back of three positive games with England in the summer, and a good pre-season, I felt really good.
"Obviously, the manager [Erik ten Hag] decided to leave me out for a game, and the team have been winning since.
"I am working hard on the training pitch to make sure I am ready for when my opportunity comes. That is all what I can do to try and help the team.
"I am not concentrating on anybody else and what people are saying, I think if people can make a story about me, I am Manchester United captain, it is going to make big news.
Maguire does not look like regaining his Manchester United place any time soon
"So, that is the reason they do it, they like the clicks, and things like that, but I went into the Euros after an eight-week injury, and did not play one game, and got into the team of the tournament.
"In terms of rustiness, and things like that, I am sure that won't happen."
Watch the World Cup with talkSPORT
At talkSPORT we are powered by fans, so come and join us for the ultimate World Cup fan experience this winter – at the talkSPORT Fan Zone.
In a huge indoor venue under the arches at Waterloo in London, we'll bring you live screenings of every World Cup game.
There will be Q&As with talkSPORT talent, you'll be part of our live broadcasts, and there will be plenty of food and drink on offer too.
Come and enjoy the best World Cup fan experience in London – and enjoy a pint on us – with tickets for all remaining games on sale HERE!
England Football


text:
Keeping the Connected Car Safe
Sentinel automotive IDS/IPS
Sentinel-ETH IDS/IPS for Automotive Ethernet
Sentinel-CAN IDS/IPS for CANbus
Sentinel-TRK IDS/IPS for SAE J1939 commercial vehicles
Sentinel-V2X IDS/IPS for V2X
Ayal Lichtblau joins Arilou Cyber Security as Vice President
November 1. 2017 Press releases
Israel-based automotive cybersecurity pioneer Arilou, member of NNG Group welcomes new Vice President to boost business development
Arilou Cyber Security, a market leader in automotive vehicle cybersecurity with a patented solution offering full protection for the car, announced the joining of Ayal Lichtblau to the company as Vice President of Sales and Business Development. He is bringing almost two decades of sales leadership experience on board, gained at companies like Samsung, cVidya Networks and Compwise. Throughout his career he held various positions at Vice President and Sales Director levels, and has obtained substantial experience in business development in Asian markets, sales management, and building up new market opportunities.
He arrives from BeGlobal, a business-consulting firm, at which he fulfilled the position of Director of Sales and Business Development, having been responsible for establishing new business partnerships between European and Korean companies in the automotive and technology sector.
As connectivity brings a plethora of great new features, the automotive industry faces a completely new kind of threat as well, and solutions are needed quickly and on a large scale. This is a truly special market that is being formed in these very moments - I find it thrilling to be part of it, and to join a company which has a pioneering role in addressing these new issues. And Arilou surely intends to maintain this role in the market. I can't wait to have our latest innovation announced shortly, that will surely cause further excitement in the industry.
He commented on his arrival.
Mr Lichtblau is a graduate of Tel Aviv University, and completed an MBA program at Bradford University.
Published bypublished
www.nng.com
©2023 - ARILOU Automotive Cybersecurity, part of NNG Group. All rights reserved.
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text:
Daniel L. Buttry
The Rev. Dr. Daniel L. Buttry is one of the world's leading peacemakers. He made history as the first full-time, global peace trainer and negotiator employed by a mainline denomination: American Baptist Churches (ABC), which counts the Rev. Dr. Martin Luther King Jr. as its most famous pastor.
He works where the world is hurting
As a peacemaker, Dr. Buttry is constantly in motion, training groups in conflict transformation, nonviolence and peace-building. List the world's hottest points of conflict and Dr. Buttry likely has worked there—from Burma/Myanmar and the former Soviet Union to areas in the Southern Hemisphere—or he has helped to train colleagues working in those venues. When he isn't traveling, Dr. Buttry is in demand to give talks and lead training events. Given his extensive experience in the world's hot spots, he also is called upon by journalists for grassroots perspective on the latest conflicts—and on connections between these hotspots and life in the U.S. For inquiries about reaching Dr. Buttry, email us at ReadTheSpirit@gmail.com
Inspiring your group
His three books for ReadTheSpirit, as of fall 2012, include the two-volume Interfaith Heroes series, profiling men and women who risked crossing religious boundaries to make peace, and the landmark Blessed Are the Peacemakers, a major book packed with inspiring peacemaker profiles from around the world. All three books have been enjoyed in individual study, small-group discussion and high school and university classes. Campus ministers and chaplains find college students especially interested in these topics.
A pastor and activist
Dr. Buttry has pastored churches in Boston and Detroit, directed the Peace Program for the ABC, and represented the ABC at the United Nations as a Non-Governmental Organizations delegate. He is a founding member of Interfaith Partners in Detroit. His wife Sharon is an ordained American Baptist clergyperson, as well, and they have three adult children. They live in Hamtramck, Mich.


As I head up the Technical and Service departments at Intergas, I guess you'd expect me to say that servicing one of our boilers is more of a pleasure than a pain. But as a heating engineer, I appreciate anything that makes servicing a boiler hassle-free, which is why I can state categorically that Intergas leads the field in this area. If you've installed and serviced an Intergas boiler, you'll know what I mean; if you haven't, prepared to be pleasantly surprised.
When the Intergas prototype was designed over 20 years ago, the focus wasn't only on delivering energy efficiency and performance, it took ease of servicing and repairs into account too. One of the biggest complaints about servicing, for example, has always been about the inaccessibility of components, like the condensate trap and heat exchanger. When parts are hard to get at, a relatively simple job turns into a time-consuming and fiddly one. In every Intergas boiler the condensate collector and trap are at the front of the boiler and are easily removed so they can be emptied and cleaned. The collector can then be clicked into place and the trap, once refilled with water, push-locked into place; if there's no reassuring sound, you know it's been incorrectly replaced and you need to do it again. The heat exchanger is also at the front of the boiler, you just have to remove the front plate by undoing the bolts, then remove and check the seal around the plate to check for any cracks. You need to use a service brush to clean the heat exchanger but generally very little cleaning is necessary because the boiler condenses in both heating and hot water modes, and any particles tend to get washed down into the collector.
Then there's the spark electrode which, in all other boilers often needs to be replaced because its long and thin gauge makes it more prone to becoming misshapen by the heat. When this happens it cannot spark or rectify correctly. Its ceramic casing is similary at risk of cracking in the heat. However, the spark electrode in an Intergas boiler rarely needs replacement because it's a much shorter and thicker component and therefore more able to withstand consistently high temperatures.
And, finally, checking and adjusting the CO2 on boilers can be problematic and time-consuming as this has to be done in 'high', 'low' and then 'high' again… but not with Intergas. The software in each boiler allows the set-up, checking and adjustments to be made in 'low' alone. Job done.


As a gift or for celebrations only the best vodka.
IMPERIA Vodka is made as a flawless vodka to the highest standards for those who appreciate the finer things in life.
Made in the world's most advanced distillation process and filtered through layers of crystal quartz from the Ural Mountains, IMPERIA is stored for 72 hours in special relaxation barrels before being bottled.
The result is an exceptionally pure and clear vodka with a light herbal aroma and an elegant, full body.


text:
Amphiura senegalensis is een slangster uit de familie Amphiuridae.

De wetenschappelijke naam van de soort werd in 1970 gepubliceerd door Fritz Jensenius Madsen.

Ophiurida


Dr. Miller's research involves the large scale simulation and modeling of turbulent air-hydrocarbon mixing and reaction at both atmospheric pressure and supercritical pressures relevant to modern and forthcoming gas turbines and diesel engines. The research is directed at fundamental studies of single-phase and multiphase flows involving complex physics in relatively simplified geometries such as homogeneous turbulence, mixing layers and jets. The preferred research approach is computational fluid dynamics (CFD); in particular, the direct numerical simulation technique. Results obtained from the simulations are used both to study the flow physics and to aid in the development and testing of mathematical and stochastic models relevant to future engineering applications. Models used in large eddy simulations (LES) and probability density function (PDF) methods are of particular interest. Direct numerical simulation, or `DNS', is an advanced computational technique in which all of the length and time scales of the governing equations are completely resolved, without resort to modeling. When conducted with high order accurate numerical algorithms, DNS therefore yields very nearly the exact solution to the problem. In this case, the results contain all of the information describing a flow at nearly all points in both time and space. This is incontrast to experiments in which information is known only where data gathering probes are installed in the flow. However, due to computational limitations DNS is a valid approach only for relatively moderate to low Reynolds number turbulence for which the range of length and time scales remains resolvable by available computer resources.
J.W. Foster and R.S. Miller, "Fundamentals of High Pressure Combustion," Chapter in High Pressure Processes in Chemical Engineering, Edited by Maximillian Lackner, ProcessEng Engineering GmbH, pp. 53-75,(2010).
J. W. Foster and Miller, R.S.,"A Priori Analysis of Subgrid Mass Diffusion Vectors in High Pressure Turbulent Hydrogen/Oxygen Reacting Shear Layer Flames," Physics of fluids, 24, 075114, (2012).
K. Ammigan, R.S., "Effect of Asymmetric Radiant Heating on Monodisperse Acetone/Ethanol and Acetone/2-Propanol Bi-Component Droplets," International Journal of Multiphase Flow, 38, 67-72, (2011).
K. Ammigan, R.S. Miller, and H.L. Clack,"Vaporization of Bi-Component Droplets Exposed to Radiant Heating," Combustion Science and Technology, 183, 1412-1432 (2011).
K. Acampora, R.S. Miller, E.M. Langan and M. LaBerge,"Development of a Novel Vascular Simulator and Injury Model to Evaluate Smooth Muscle Cell Response Flowing Balloon Angioplasty," The Annals of Vascular Surgery, 21 (6) 734-741,(2007).


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Health & Physical Education at University of Michigan - Ann Arbor
270 Bachelor's Degrees
46 Master's Degrees
3 Doctor's Degrees
The Health & Physical Education major is part of the parks, recreation, leisure, & fitness studies program at University of Michigan - Ann Arbor.
Here, you'll find out more about the major, including such details as the number of graduates, what degree levels are offered, ethnicity of students, average salaries, and more. In addition, we cover how U-M ranks in comparison to other schools with phys ed programs.
Go directly to any of the following sections:
Health & Physical Education Degrees Available at U-M
Bachelor's Degree in Phys Ed
Master's Degree in Phys Ed
Doctorate Degree in Phys Ed
U-M Phys Ed Rankings
Ranked at #3 in College Factual's most recent rankings, U-M is in the top 1% of the country for phys ed students pursuing a bachelor's degree. It is also ranked #1 in Michigan.
Best Health & Physical Education Bachelor's Degree Schools 1
Best Health & Physical Education Doctor's Degree Schools 3
Best Health & Physical Education Master's Degree Schools 5
Most Popular Doctor's Degree Colleges for Health & Physical Education 15
Most Popular Bachelor's Degree Colleges for Health & Physical Education 28
Most Focused Doctor's Degree Colleges for Health & Physical Education 32
Most Popular Master's Degree Colleges for Health & Physical Education 36
Best Value Doctor's Degree Colleges for Health & Physical Education 44
Most Focused Master's Degree Colleges for Health & Physical Education 228
Best Value Master's Degree Colleges for Health & Physical Education 282
Best Value Bachelor's Degree Colleges for Health & Physical Education 501
Most Focused Bachelor's Degree Colleges for Health & Physical Education 627
Popularity of Phys Ed at U-M
During the 2019-2020 academic year, University of Michigan - Ann Arbor handed out 270 bachelor's degrees in health and physical education. Due to this, the school was ranked #28 in popularity out of all colleges and universities that offer this degree. This is an increase of 13% over the previous year when 240 degrees were handed out.
In 2020, 46 students received their master's degree in phys ed from U-M. This makes it the #34 most popular school for phys ed master's degree candidates in the country.
In addition, 3 students received their doctoral degrees in phys ed in 2020, making the school the #36 most popular school in the United States for this category of students.
How Much Do Phys Ed Graduates from U-M Make?
$39,000 Bachelor's Median Salary
Salary of Phys Ed Graduates with a Bachelor's Degree
Phys Ed majors who earn their bachelor's degree from U-M go on to jobs where they make a median salary of $39,000 a year. This is great news for graduates of the program, since this figure is higher than the national average of $28,000 for all phys ed bachelor's degree recipients.
How Much Student Debt Do Phys Ed Graduates from U-M Have?
$16,848 Bachelor's Median Debt
Student Debt of Phys Ed Graduates with a Bachelor's Degree
While getting their bachelor's degree at U-M, phys ed students borrow a median amount of $16,848 in student loans. This is not too bad considering that the median debt load of all phys ed bachelor's degree recipients across the country is $25,000.
The typical student loan payment of a bachelor's degree student from the phys ed program at U-M is $175 per month.
Phys Ed Student Diversity at U-M
Take a look at the following statistics related to the make-up of the phys ed majors at University of Michigan - Ann Arbor.
Is U-M a Good Match for You?
U-M Health & Physical Education Bachelor's Program
Of the 270 students who earned a bachelor's degree in Health & Physical Education from U-M in 2019-2020, 47% were men and 53% were women.
The majority of the students with this major are white. About 75% of 2020 graduates were in this category.
The following table and chart show the ethnic background for students who recently graduated from University of Michigan - Ann Arbor with a bachelor's in phys ed.
Hispanic or Latino 18
Other Races 19
U-M Health & Physical Education Master's Program
For the most recent academic year available, 60% of phys ed master's degrees went to men and 40% went to women.
The following table and chart show the ethnic background for students who recently graduated from University of Michigan - Ann Arbor with a master's in phys ed.
Non-Resident Aliens 14
Image Credit: By Dwight Burdette under License


text:
USC School of Law may refer to:

 University of South Carolina School of Law
 University of Southern California Law School


What's up Don, one of my favorite people!!!!
@TTCavegn TIff!! Miss ya, friend.
@Alison_Wedig do you want to ask your question on screen?
Communication and listening, this is good information! Thanks for sharing.
Dang, Ian is a master!
Just go all in Joel!
Dont forget to send me your contact info!!!
@jondercin way to go Joel!
Thanks. I have to run in about 5 minutes. I'll stick around in case I can help.
Thanks @3rhinomedia and @GeorgeBThomas ! I'm out!
Ian – we need to talk about an online course!!!!
let's get ian to 1000 hands!


Diving is a wonderful experience and beautiful art to learn. Though open water scuba diving within 30 meters is great fun, it is just not deep enough to discover bigger and deeper dive sites around the world. As adventurous divers you will be eager to go a little deeper and have more fun. These and more are the things you get to learn from PADI Deep Diver Specialty course.
The PADI Deep Diver Specialty course allows you to safely explore beyond open water diving training limits. Exciting and thrilling experience such as shipwrecks, caves, and beautiful wall dives lay beyond open water diving.
The purpose of Pisces PADI Deep diver specialty course is to serve as a safe, supervised introduction to deeper diving within the limits of recreational scuba diving with in-depth emphasis on fun and safety.
For deep dives, the minimum depth is between 18-30 meters and them maximum diving depth is 40 meters with all dives conducted within the no-decompression limits. The Deep Diver Specialty course equips students with the theoretical knowledge and practical experience of deep diver. At the end of the course, the student would be able to plan, organize and confidently dive to depths of between 18 and 40 meters.
With considerations such as safety stops, decompression-table reviews, and special equipment requirements this course will give you the perfect techniques and confidence to explore deeper.
The course is applicable for any certified open water scuba diver who would love to expand their diving knowledge and expertise. PADI Deep diver Specialty course is an excellent option for certified open water scuba divers that seek to advance their dive training.
To enroll for the Deep diver specialty course, the student must be 18 years, but a minimum of 15 years with parental consent is acceptable.
Applicant must be a certified PADI Adventure Diver or have attained equivalent qualification.
PADI's Deep Diver study material will provide a thorough overview of the techniques and principle needed to complete the PADI Deep Diver specialty course. Also, all required equipment for the duration of the course will be at your disposal.
Plan, organize and make four deep dives under instructor's supervision. You will be taught decompression illness, review effects of nitrogen narcosis and other symptoms and deep water diving first aid techniques.
It takes two days to complete the course where all principal evaluation and requirement by PADI will be completed. Students will complete four dives at a maximum depth of 40 meters within two days.
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I was watching tv (comedy movie) and relaxing on my couch. All of a sudden I had a wave of lightheadedness and foggyish feeling then my body felt very hot and heart started pounding. Took a Ativan and fell asleep shortly.
Woke up in the morning confused like why did this happen so randomly.
It's all anxiety, it can't harm you if you don't let it get to you.
I have experienced the same wave of panic you described but it usually happens to me in public.


Mosaic House takes pride in its commitment to sustainability. The hotel has been redesigned to feature solar panels on its roof, ensuring that 100 per cent of its electricity is generated by renewable sources. Intelligent heating and window and facade insulation, energy-efficient lighting, electric shutters, and a grey-water system that recycles shower and sink water for toilets make the Mosaic House a truly unique green property.The hotel offers a central location, cosy private rooms with free Wi-Fi and air-conditioning, shared rooms for a more social environment, and luxurious design apartments with fantastic amenities. Each one has a stylish, modern style. There is also a restaurant, music bar, and lounge, offer fresh breakfast in the morning, and music and entertainment at night.
Burgers, salads, sandwiches and more are on offer at the Mosaic House on-site restaurant.
Just a few minutes away from the main sights of Prague, Mosaic House hotel is ideally located for business and leisure travellers to the Czech capital. Important attractions like the Charles Bridge, Old Town Square, and the John Lennon Wall are all easy to reach from the hotel. Prague Castle, the Great Strahov Stadium and the GENERALI Arena are all further away. Prague central railway station is easily accessible.