text:
First Quantum ships first copper concentrate from Cobre Panama mine
Cecilia Jamasmie | June 17, 2019 | 10:29 am Base Metals Canada Latin America Copper
(Image courtesy of First Quantum)
Canadian miner First Quantum Minerals (TSX:FM) has made the first shipment of concentrate from its massive mining and processing complex in Panama, the largest copper operation coming to market over the next couple of years and the biggest single private sector investment in the nation's history.
The key milestone for Cobre Panama, located about 120 km west of Panama City and 20 km from the Atlantic coast, comes at a time when the market is in need of high concentrate volumes, due to recent supply disruptions, including a strike in the world's largest producer, Chile's Codelco.
Cobre Panama processed its first ore in early February, fifty years after the deposit was discovered, but only six years since First Quantum gained control over it
The project is on track to deliver between 140,000 and 175,000 tonnes in 2019. About 80% of total output will be generated in the second half of the year, as the mill ramps up to its annual capacity of 72 million tonnes by the end of this year.
After that, the Toronto-based company expects to increase copper output from 85 million tonnes per year to 100 million annual tonnes, beginning in 2023.
The expanded throughput involves the earlier development of the adjacent Colina pit, the addition of a ninth mill, an expanded mining equipment fleet, additional conveyors, an in-pit crusher and other infrastructure to access the nearby pit.
In 2017, First Quantum spent close to $1 billion to advance construction at Cobre Panama (cobre is Spanish for copper), investing a further $830 million in 2018. This year, the company plans to allocate around $110 million to keep advancing the project until it reaches its full capacity of more than 375,000 annual tonnes of copper.
Cobre Panama processed its first ore in early February, fifty years after the deposit was discovered, but only six years since First Quantum gained control over it through the acquisition of rival Canadian copper miner Inmet Mining.
Once the mine reaches full production, expected in 2021, First Quantum's total production will surpass 900,000 tonnes a year, placing it among the world's top six copper producers.
Panama will also reap the benefit from the operating open-pit mine, as it is expected to generate around $2 billion worth of annual exports during its 34-years of life. That, according to Christie, is equivalent to around 4% of the Central American nation's current GDP.
First Quantum, which has operating mines in Australia, Zambia, Mauritania, Turkey, Spain and Finland, is also developing two other projects in Latin American — Haquira in Peru and Taca Taca in Argentina – but the miner hasn't decided which one it will develop first.
Western Potash begins production drilling at Milestone
This is the world's first potash mine that will leave no salt tailings on the surface.


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Microsoft releases feature-complete Hyper-V code
Windows Server 2008 Revealed: Hyper-V virtualization
Review: Much to like in Windows Server 2008
Ballmer launches Windows Server 2008, lauds user base
New Windows Server will lead march to 64-bit OS
By James Niccolai and Elizabeth Montalbano
The launch of a new family of Windows server products this week will kick-start a broad shift among customers to 64-bit versions of Microsoft's server software, analysts and customers said.
Microsoft Corp. CEO Steve Ballmer is due to launch two major product upgrades at an event Wednesday in Los Angeles -- the Windows Server 2008 operating system, which is due for release next week, and its SQL Server 2008 database, expected in the third quarter, following delays. He's also expected to discuss Visual Studio 2008, which shipped in November.
Like their predecessors, the new products will be offered in both 32- and 64-bit editions. But several factors this time will prompt more customers to choose the 64-bit versions, including the broad availability of 64-bit, x86-based server hardware and the trend toward consolidating and virtualizing server workloads to reduce power consumption and improve efficiency.
The shift will happen gradually, since most customers are not expected to deploy the products widely until next year. But it will mark a significant maturation for Microsoft's server products, which long were seen as an also-ran in the data center beside 64-bit Unix operating systems from companies such as Sun Microsystems Inc. and Hewlett-Packard Co. It should also mean better performance for Microsoft customers.
"This will absolutely tip the scales in terms of more 64-bit deployments moving forward," Gartner Inc. analyst John Enck said of the new products. The move will be driven by a desire among customers to get the most out of their 64-bit server hardware, he said, which means using a 64-bit operating system.
The difference lies in the amount of physical memory the software can address. A 32-bit OS can address only 4GB of main memory without having to use technology tricks that diminish performance gains. A 64-bit OS can address far more memory -- up to 2TB in the case of Windows Server 2008, according to Microsoft.
That will boost the performance of some applications, because they will be able to pull data quickly from main memory instead of having to retrieve it from disk, which is slower. The gains should be evident for databases and for Microsoft's Exchange Server, although line-of-business applications will see less benefit, Enck said.
Customers may also be driven to 64 bits by concerns about the future. Microsoft has said this will be the last big upgrade to Windows Server offered in both 32- and 64-bit editions, and some expect the same to be true for SQL Server. Exchange Server 2007, released in November, already is available only in 64 bits. Customers would be wise to start preparing now for a move that soon will be forced upon them anyway, analysts said.
Also propelling the move is the trend toward server consolidation. Virtualization allows multiple operating systems and application loads to run on a single physical machine; server virtualization requires the capacity of a powerful, 64-bit server.
"Anybody doing a deployment today would be foolish not to at least consider when and where a 64-bit OS would be a good fit," said Al Gillen, an analyst IDC. "It's really about future-proofing your IT environment -- giving yourself the ability to support the workloads that you'll have on these servers before they are retired in five years' time."
IDC has called the lack of adoption of 64-bit Windows Server "one of the biggest missed opportunities among today's customer base." It has noted that the 64-bit products will be priced the same as their 32-bit counterparts and argued that the transition is relatively easy for customers.
The 64-bit processors from Intel Corp. and Advanced Micro Devices Inc. have maintained the x86 architecture from the 32-bit world. That means 32-bit applications can still run on 64-bit servers and that "the majority of existing 32-bit applications will run aboard 64-bit Windows Server without modification and, most frequently, with improved performance," IDC said.
Customers will need to update low-level system tools such as security products, antivirus tools and some system-management products, which interact directly with the Windows Server kernel, IDC said.
At the end of 2007, the research firm estimates, only about 10% of Windows Server customers were using the 64-bit edition of Windows Server 2003. It expects that figure to approach 50% by the end of 2010, driven by Windows Server 2008. For new licenses sold in 2010, close to 75% will be for a 64-bit version of Windows Server, according to IDC.
Microsoft is keen to promote the transition to 64 bits. The technology will give its customers better performance and help the company catch up with the Unix world, where powerful -- but more expensive -- servers from Sun, IBM and HP have long been based on 64-bit operating systems. Microsoft believes that the products launching this week will mark "a big turning point" in the use of 64-bit Windows software, said Ward Ralston, a Microsoft senior technical product manager.
Microsoft also is pressuring independent software vendors to get their software 64-bit ready, after a lack of preparedness held back the transition following Windows Server 2003's release. Such vendors aren't required to have a native 64-bit edition of their software to receive a "Certified for Windows Server 2008" logo, but they will need to ensure that their software can run on the 64-bit OS.
Some expect Microsoft's virtualization technology, Hyper-V, to be a factor. The hypervisor will be offered free with the 64-bit edition of Windows Server 2008. Andrew Brust, head of new technology at the IT consulting company Twentysix New York, a Microsoft partner, said that once Hyper-V ships, it's "going to be huge. The virtualization space is ripe for some new competition."
Brian Randell, a senior consultant at another Microsoft partner, MCW Technologies in Los Angeles, said Hyper-V will be a major impetus for the move to 64 bits. "It demands that you have that kind of processor environment available," he said.
However, others pointed to Hyper-V's immaturity. It was originally planned to ship with Windows Server 2008 but has been delayed for up to six months. Even then it will be Microsoft's first attempt at virtualization, noted Michael Cherry, an analyst at Directions on Microsoft.
"I think there's too much emphasis on virtualization with this release," he said. Hyper-V may eventually play a significant role for Microsoft, but the company first needs to develop the required tools for managing a complex virtualized environment. After the hypervisor is released, Cherry said, Microsoft will also need to update its Virtual Machine Manager product.
Still, Cherry is upbeat about the new products, particularly Windows Server 2008. The redesign of the operating system to allow customers to install only the functions they need for particular tasks, or roles, will provide security and maintenance advantages, he said. He also pointed to the new Internet Information Server, which gives more options for running and controlling applications remotely, and a significant update to Terminal Services, which will make it easier to run line-of-business applications on a server and make them appear to the end user as if they were running locally.
Brust said the new server products are "rock-solid on 64-bit, and so too are the currently shipping versions of SQL Server, SharePoint and other server applications."
"When SQL Server 2008 ships, it will be the third version of the product to offer 64-bit support," Brust said. "Let's face it -- it's time to move off 32-bit."

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}\label{sec_intro}

For a $k$-graph $\GG=(V,E)$, i.e.\ a $k$-uniform hypergraph on a vertex set $V$ and an edge set $E\subseteq \binom{V}{k}$,
a \textit{$2$-edge coloring} of $\GG$ is a function $c\colon E(\GG)\to \{\text{red},\text{blue}\}$
that maps each edge to one of the given colors \textit{red} or \textit{blue}.
In the following refer to such a function simply as a \textit{coloring} of $\GG$.
We say that a $k$-graph $\GG$ has the \textit{Ramsey property} $\GG\rightarrow \HH$ for some $k$-graph $\HH$ if each coloring of $\GG$ contains a monochromatic copy of $\HH$.
The \textit{size-Ramsey number} of a $k$-graph $\HH$ is defined as
$$\RRR^{(k)}(\HH)=\min\big\{m\in\N \colon \exists\ k\text{-graph }\GG\text{ with }|E(\GG)|=m\text{ and } \GG\rightarrow \HH\big\}.$$

Size-Ramsey problems have been firstly introduced by Erd\H{o}s, Faudree, Rousseau and Schelp \cite{erdos_size} for graphs.
One of the focus points of studies on the graph case is estimating the size-Ramsey number of paths.
Beck \cite{beck83} disproved a conjecture of Erd\H{o}s \cite{erdos_pn} by showing that $\RRR^{(2)}(P_n)=O(n)$.
Since then, estimates on this number have been gradually improved, with the current best known bounds being
$\big(3.75-o(1)\big)n\le\RRR^{(2)}(P_n)\le74n$
given by Bal, DeBiasio \cite{bal_lb} and Dudek, Pralat \cite{dud_74}, respectively.
\\

Let $n,k,\ell\in\N$ with $k\ge2$, $1\le \ell \le k-1$ and $\frac{n-\ell}{k-\ell}\in\N$.
A \textit{$k$-uniform $\ell$-path} on $n$ vertices $\PP^{(k)}_{n,\ell}$ is a $k$-graph on $n$ vertices
for which there exists an ordering of its vertices such that every edge is a $k$-element set of consecutive vertices (i.e.\ an interval of length $k$),
two consecutive edges have precisely $\ell$ vertices in common, and there are no isolated vertices.
A $k$-uniform $\ell$-path with $\ell=k-1$ is referred to as a \textit{$k$-uniform tight path}.
If the uniformity is given from the context we omit the term `$k$-uniform'.
\\

Research on the size-Ramsey number of hypergraphs has been substantially driven forward by Dudek, La Fleur, Mubayi and Rödl \cite{dud_general}.
Among other results, they conjectured that the size-Ramsey number of tight paths is linear in terms of $n$.
This conjecture was lately verified by Letzter, Pokrovskiy and Yepremyan \cite{let_ub}.

\begin{theorem}[\cite{let_ub}]\label{ub_linear}
Let $k\ge 2$ be fixed. Then
$$\RRR^{(k)}(\PP^{(k)}_{n,k-1})=O(n).$$
\end{theorem}

Regarding a lower bound on this number, it is a simple observation that the following trivial bound holds.

\begin{observation*}\label{lb_trivial}
Let $n,k,\ell\in\N$, $k\ge2$, $1\le \ell\le k-1$ with $\tfrac{n-\ell}{k-\ell}\in\N$. Then
$$\RRR^{(k)}(\PP^{(k)}_{n,\ell})\ge \frac{2(n-\ell)}{k-\ell}-1.$$
In particular, $\RRR^{(k)}(\PP^{(k)}_{n,k-1})\ge 2n-2k+1$ for any $n,k\in\N$, $k\ge2$.
\end{observation*}
\medskip

In this paper we show an improved lower bound on the size-Ramsey number of tight~paths.

\begin{theorem}\label{lb_tight2}
Let $n\ge 7$. Then
$$\RRR^{(3)}(\PP^{(3)}_{n,2})\ge\tfrac{8}{3} n -\tfrac{28}{3}.$$
\end{theorem}

\begin{theorem}\label{lb_tight}
Let $k\ge4$ and $n>\frac{k^2+k-2}{2}$. Then
$$\RRR^{(k)}(\PP^{(k)}_{n,k-1})\ge \big\lceil\log_2(k+1)\big\rceil\cdot n-2k^2.$$
\end{theorem}
\medskip

Section \ref{sec_neigh} discusses useful properties for the main proofs.
In Section \ref{sec_tight} the proofs of Theorem \ref{lb_tight} and Theorem \ref{lb_tight2} are presented.
Lastly, Section \ref{sec_con} of this paper briefly describes how the mentioned proofs can be applied to obtain a lower bound for general $k$-uniform $\ell$-paths.
\bigskip

\section{Preliminaries}\label{sec_neigh}

Let $\GG$ be a $k$-graph and $Z\subseteq E(\GG)$ be an edge set.
Let $\medcup Z =\{v\in e\colon e\in Z\}$ be the set of vertices that are \textit{covered} by~$Z$.
We say that the $k$-graph $(\medcup Z, Z)$ is \textit{formed} by $Z$.
Given a vertex set $W\subseteq V(\GG)$ the subhypergraph \textit{induced} by $W$ is $\GG[W]=(W,\{e\in E(\GG)\colon e\subseteq W\})$.
For $q\in\R$, $0\le q< k$, the \textit{$q$-neighborhood} of $Z$ is the edge set
$$N_{> q}(Z)=\big\{e\in E(\GG) \colon \exists e'\in Z\text{ with }|e\cap e'|> q\big\}.$$
Note that we allow $e=e'$, thus $Z\subseteq N_{>q}(Z)$ for all $0\le q< k$.
\\

For each $k$-uniform $\ell$-path $\PP$ on $n$ vertices we fix an ordering of the vertices such that each edge is a set of consecutive vertices.
We say that such an enumeration $V(\PP)=\{v_1,\dots,v_n\}$ is \textit{according to} $\PP$.
For a $k$-graph $\GG$, we define $e(\GG)=|E(\GG)|$.
Observe that $e(\PP^{(k)}_{n,\ell})=\frac{n-\ell}{k-\ell}$ and in the tight case $e(\PP^{(k)}_{n,k-1})=n-k+1$.
Furthermore, let $[n]=\{1,\dots,n\}$ for $n\in\N$. For any other notation, see Diestel \cite{diestel}.

\begin{proposition}\label{prop_gen}
Let $n,k\in\N$, $n\ge k\ge 2$ and let $\PP$ be a $k$-uniform tight path on $n$ vertices.
Furthermore, let $\alpha\in\R$, $1\le \alpha \le k$ be a constant and $W\subseteq V(\PP)$ be a vertex set such that for each edge $e\in E(\PP)$ we have $|e\cap W|\ge \alpha$.
Then $$|W|\ge \frac{\alpha(n-k+1)}{k}.$$

\noindent In particular, if for each $e\in E(\PP)$, $|e\cap W|> \frac{k+1}{2}$, then for $n>\frac{k^2+k-2}{2}$, $$|W|> \frac{n}{2}.$$
\end{proposition}

\begin{proof}
We estimate the size of $W$ by double-counting ordered pairs $(v,e)$ consisting of a vertex $v\in W$ and an edge $e\in E(\PP)$ with $v\in e$.
Let $\rho_{(v,e)}$ be the number of such pairs.

Considering the edges of $\PP$ it is immediate that
$$\rho_{(v,e)}\ge\alpha \cdot e(\PP)= \alpha (n-k+1).$$

Now consider the vertices in $W\subseteq V(\PP)$.
The maximum degree of the tight path $\PP$ is at most $k$, so
$$\rho_{(v,e)}\le k\cdot |W|.$$

\noindent Combining both inequations, we obtain
$$|W|\ge \frac{\alpha (n-k+1)}{k}.$$

Now consider the case that for each edge $e\in E(\PP)$ we have $|e\cap W|>\frac{k+1}{2}$, then also $|e\cap W|\ge\frac{k+2}{2}$.
Therefore we obtain for sufficiently large $n$, $$|W|\ge \frac{k+2}{2}\cdot\frac{n-k+1}{k}> \frac{n}{2}.$$
\end{proof}
\smallskip

\section{Proofs of the main results}\label{sec_tight}

\begin{proof}[Proof of Theorem \ref{lb_tight}]
Let $\GG$ be a $k$-uniform hypergraph with $\GG\rightarrow \PP^{(k)}_{n,k-1}$, i.e.\ such that each coloring contains a monochromatic $k$-uniform tight path on $n$ vertices.
We show that there are at least $\left\lceil\log_2(k+1)\right\rceil\cdot n-2k^2$ many edges in $\GG$
by iteratively constructing many edge-disjoint tight paths.
Let $\lambda=\left\lceil\log_2(k+1)\right\rceil-1$, this number indicates how many iteration steps are executed.
Additionally, we define the function $q\colon\{0,\dots,\lambda\}\rightarrow \R$,
$$q(i)=\left(1-\frac{1}{2^i}\right)(k+1),$$
which will be the parameter of the $q$-neighborhoods considered in each iteration step.
Clearly, $q$ is an increasing function and $q(i)\ge 0$ for $i\in\{0,\dots,\lambda\}$.
For $i\le\lambda$ (or equivalently $i<\log_2(k+1)$) it can be seen that $q(i)<k$,
which implies that the $q(i)$-neighborhood is well-defined for all $i\in\{0,\dots,\lambda\}$.
\\

As an initial step of the iteration, the Ramsey property $\GG\rightarrow \PP^{(k)}_{n,k-1}$ provides that there is some tight path on $n$ vertices in $\GG$, which we denote by $\PP_0$.
\\

\noindent From now on we proceed iteratively, so let $i= 1,\dots,\lambda$ and suppose that the iteration has been performed for all smaller values of $i$.
In each step of the iteration we construct the following:
\begin{itemize}
\item Edge sets $Z^1_i, Z^2_i\subseteq E(\PP_{i-1})$ such that $\medcup Z^1_i \cap \medcup Z^2_i=\varnothing$ and each of the sets forms a tight path in $\GG$ on precisely $\left\lfloor\frac{n}{2}\right\rfloor$ vertices.

\item A tight path $\PP_{i}$ on $n$ vertices with $E(\PP_{i})\cap N_{>q(i)}(Z^{a}_{b})=\varnothing$ for all $a\in[2], b\in[i]$.
\end{itemize}

\noindent As a first step, we construct $Z^1_i$ and $Z^2_i$ by dividing the tight path $\PP_{i-1}$ into two parts of equal length
and considering the edge sets of the two created shorter tight paths.
For this purpose, consider an ordering of the vertices $V(\PP_{i-1})=\{v_1,\dots,v_n\}$ according to $\PP_{i-1}$.
Let $$V^1_i=\big\{v_1,\dots,v_{\left\lfloor\frac{n}{2}\right\rfloor}\big\}\quad\text{ and }\quad V^2_i=\big\{v_{\left\lceil\frac{n}{2}\right\rceil+1},\dots,v_n\big\}.$$
Then $|V^1_i|=\left\lfloor\frac{n}{2}\right\rfloor=|V^2_i|$.
Now let $Z^1_i=E(\PP_{i-1}[V^1_i])$ and $Z^2_i=E(\PP_{i-1}[V^2_i])$. Clearly, these two sets form vertex-disjoint tight paths on $\left\lfloor\frac{n}{2}\right\rfloor$ vertices in $\GG$.
The size of $Z^1_i$ and $Z^2_i$ is
$$|Z^1_i|=|Z^2_i|=e(\PP^{(k)}_{\left\lfloor\tfrac{n}{2}\right\rfloor,k-1})=\left\lfloor\frac{n}{2}\right\rfloor-k+1\ge \frac{n-2k+1}{2}.$$

\noindent
In a next step we show a key property of the edge sets $Z^{a}_{b}$ for $a\in[2]$, $b\in[i]$.
\\

\noindent \textbf{Claim.} Let $a_1,a_2\in[2]$, $b_1,b_2\in[i]$ such that $(a_1,b_1)\neq (a_2,b_2)$.
Then for any two edges $e_1\in N_{>q(i)}(Z^{a_1}_{b_1})$ and $e_2\in N_{>q(i)}(Z^{a_2}_{b_2})$ we have $$|e_1\cap e_2|<k-1.$$
\smallskip
\noindent \textit{Proof of the claim.}
Assume that there are edges $e_1\in N_{>q(i)}(Z^{a_1}_{b_1})$, $e_2\in N_{>q(i)}(Z^{a_2}_{b_2})$ with $|e_1\cap e_2|\ge k-1$.
By definition, there is an edge $z_1\in Z^{a_1}_{b_1}$ such that $|e_1\cap z_1|>q(i)$ and an edge $z_2\in Z^{a_2}_{b_2}$ with $|e_2\cap z_2|>q(i)$.
\\

\begin{figure}[H]
\centering
\includegraphics[scale=0.65]{figs/neigh}
\caption{Possible constellation of the edges in iteration step $i=1$ where $k=6$}
\end{figure}

\noindent We estimate the size of $z_1\cap z_2$ in order to find a contradiction to our assumption.
Since $|e_1\cap e_2|\ge k-1$, we have $|e_1\backslash e_2|\le 1$ and so $|e_2\cap z_1|>q(i)-1$.
Applying this, we obtain:
\begin{multline*}
|z_1\cap z_2|\ge |e_2\cap z_1\cap z_2|\ge |e_2|-|e_2\backslash z_1|-|e_2\backslash z_2|= -|e_2|+|e_2\cap z_1|+|e_2\cap z_2|\\
>-k+q(i)-1+q(i)=\left(1-\frac{1}{2^{i-1}}\right)(k+1)=q(i-1).
\end{multline*}

If $b_1=b_2$, we have $\medcup Z^{a_1}_{b_1}\cap \medcup Z^{a_2}_{b_2}=\varnothing$ by construction. Then $|z_1\cap z_2|=0\le q(i-1)$, which is a contradiction.

We suppose that $b_1\neq b_2$, then without loss of generality $b_1>b_2$ (and by this $b_1-1\ge1$).
By construction we know $z_1\in Z^{a_1}_{b_1}\subseteq E(\PP_{b_1-1})$.
In the iteration step $b_1-1$ the tight path $\PP_{b_1-1}$ was chosen to be edge-disjoint from $\bigcup_{a\in[2],b<{b_1}} N_{>q(b_1-1)}(Z^a_b)$.
This yields that $z_1\notin N_{>q(b_1-1)}(Z^{a_2}_{b_2})$ and so
$$|z_1\cap z_2|\le q(b_1-1)\le q(i-1),$$
since $q$ is an increasing function, and we again reach a contradiction.
This concludes the proof of the claim. \\

Now we find the next tight path $\PP_{i}$ in $\GG$ by considering the following coloring of $\GG$.
For all $a\in[2]$ and $b\in[i]$, assign the color red to each edge in $N_{>q(i)}(Z^{a}_{b})$. The remaining edges are colored blue.
We will prove that there is a monochromatic blue $\PP^{(k)}_{n,k-1}$ in this coloring. We shall let $\PP_{i}$ be that path.
With this in mind, assume for a contradiction that there is a monochromatic red tight path $\RR$ on $n$ vertices in $\GG$.
\\

Clearly, each edge in $E(\RR)$ is in some neighborhood $N_{>q(i)}(Z^{a}_{b})$, $a\in[2], b\in[i]$.
Now the above claim provides that any two edges which are consecutive in $\RR$, so intersect in precisely $k-1$ vertices,
belong to the same neighborhood $N_{>q(i)}(Z^{a}_{b})$ for some $a\in[2], b\in[i]$.
By repeating this argument, we obtain that $E(\RR)\subseteq N_{>q(i)}(Z^a_b)$.
This implies that for all $e\in E(\RR)$, $$|e\cap \medcup Z^a_b|>q(i)\ge q(1)=\tfrac{k+1}{2}.$$
Then applying Proposition \ref{prop_gen} for the tight path $\RR$ and the vertex set $\medcup Z^a_b$ yields $|\medcup Z^a_b|>\frac{n}{2}$.
But by construction $Z^a_b$ forms a $k$-graph on precisely $\left\lfloor\frac{n}{2}\right\rfloor$ vertices, a contradiction.
\\

Consequently, there is no red tight path on $n$ vertices in the coloring,
so the Ramsey property $\GG\rightarrow \PP^{(k)}_{n,k-1}$ implies the existence of a monochromatic blue $\PP^{(k)}_{n,k-1}$, which we denote $\PP_{i}$.
Observe that for each $e\in E(\PP_{i})$, $e\notin N_{>q(i)}(Z^a_b)$ for all $a\in[2], b\in[i]$, since all edges in these neighborhoods are colored in red.
\\

By iterating the described procedure for $i=1,\dots,\lambda$, we obtain edge sets $Z^a_b$ for $a\in[2]$, $b\in[\lambda]$ which are pairwise disjoint
and additionally a tight path $\PP_{\lambda}$ on $n$ vertices such that each edge in $E(\PP_{\lambda})$ is not contained in any set $Z^a_b$.
This allows for the following estimate on the number of edges in $\GG$
$$e(\GG)\ge \sum_{b\in[\lambda]} \big(|Z^1_b|+|Z^2_b|\big) + e(\PP_{\lambda})\ge\lambda (n-2k-1)+(n-k+1)\ge \left\lceil\log_2(k+1)\right\rceil\cdot n-2k^2.$$
\end{proof}
\bigskip

We point out that the above proof also applies to $3$-uniform tight paths, but does not yield an improvement of the trivial bound.
In order to obtain a refined bound in this case, we instead use a non-iterative adaption of the above proof.
\\

\begin{proof}[Proof of Theorem \ref{lb_tight2}]
Let $\GG$ be an arbitrary $3$-uniform hypergraph which has the Ramsey property $\GG\rightarrow \PP^{(3)}_{n,2}$.
As before, we show that $\GG$ is a $3$-graph on at least $\tfrac{8}{3} n -\tfrac{28}{3}$ many edges.
Using the Ramsey property $\GG\rightarrow \PP^{(3)}_{n,2}$, there exists some tight path on $n$ vertices in~$\GG$.
In particular, we find a shorter tight path $\PP_0$ on only $n_0\coloneqq\left\lceil\tfrac{2}{3} n -\tfrac{4}{3}\right\rceil-1$ many vertices.
Observe that $e(\PP_0)= n_0 -2\ge \tfrac{2}{3}n-\tfrac{13}{3}$.
\\

In order to find a tight path $\PP_1$ which is edge-disjoint from $\PP_0$, we consider the following coloring.
Color all edges in the $1$-neighborhood $N_{>1}\big(E(\PP_0)\big)$ in red and the remaining edges in blue.
Assume for a contradiction that in this coloring there is a monochromatic red tight path on $n$ vertices, say $\RR$.
Then Proposition \ref{prop_gen} applied to the tight path $\RR$ and the vertex set $V(\PP_0)$ provides a contradiction.
Since $\GG\rightarrow \PP^{(3)}_{n,2}$, there is a monochromatic blue tight path on $n$ vertices in~$\GG$.
This implies that there is also a blue tight path on $n_1\coloneqq n-1$ vertices and by this $n-3$ edges.
We fix such a tight path $\PP_1$ with $e(\PP_1)=n-3$. Note that $N_{>1}\big(E(\PP_0)\big)$ and $E(\PP_1)$ are disjoint edge sets.
\\

In the following, in order to find a third edge-disjoint tight path, we consider another coloring of $\GG$.
From now on, let each edge in $E(\PP_0) \cup E(\PP_1)$ be colored red and all other edges blue.
Assume for a contradiction that there is a red tight path $\RR$ on $n$ vertices in this coloring.
Then neither $E(\RR)\subseteq E(\PP_0)$ nor $E(\RR)\subseteq E(\PP_1)$ since the two edge sets have a size strictly less than $e(\PP^{(3)}_{n,2})$.
Therefore, $\RR$ consists of edges of both $E(\PP_0)$ and $E(\PP_1)$.
Both of these edge sets are disjoint, so there exist two edges $e_1\in E(\PP_0)\cap E(\RR),e_2\in E(\PP_1)\cap E(\RR)$ which are consecutive in $\RR$, i.e.\ $|e_1\cap e_2|=2$.
But that is a contradiction to the fact that $N_{>1}\big(E(\PP_0)\big)$ and $E(\PP_1)$ are disjoint.
Consequently, there is no red $\PP^{(3)}_{n,2}$ in this coloring.
By the same argument as before, there is a blue tight path $\PP_2$ on $n_2\coloneqq n$ vertices in $\GG$.

Then the three edge sets $E(\PP_0), E(\PP_1), E(\PP_2)$ are pairwise disjoint. Thus,
$$e(\GG)\ge e(\PP_0)+e(\PP_1)+e(\PP_2)\ge \tfrac{8}{3} n -\tfrac{28}{3}.$$
\end{proof}

\section{Generalization to $k$-uniform $\ell$-paths}\label{sec_con}

In order to obtain a lower bound for all $k$-uniform $\ell$-paths, both proofs presented in the previous section can be generalized.

\begin{theorem}\label{lb_gen_small}
Let $k\ge2$ and $1\le\ell\le\frac{2}{3}k$. Let $n\in\N$ be sufficiently large such that $\tfrac{n-\ell}{k-\ell}\in\N$. Then
$$\RRR^{(k)}(\PP^{(k)}_{n,\ell})\ge \left(2+\frac{\ell}{k+c}\right)\cdot \frac{n-\ell}{k-\ell}-4,$$
where $c$ is the smallest non-negative integer with $\frac{k+c}{k-\ell}\in\N$.
\end{theorem}

\begin{theorem}\label{lb_gen_large}
Let $k\ge 4$ and $\frac{2}{3}k<\ell\le k-1$. Let $n\in\N$ be sufficiently large with $\tfrac{n-\ell}{k-\ell}\in\N$. Then
$$\RRR^{(k)}(\PP^{(k)}_{n,\ell})\ge \left\lceil\log_2\left(\frac{2k-\ell}{k-\ell}\right)\right\rceil \cdot \frac{n-\ell}{k-\ell} - 4k^2.$$
\end{theorem}

\noindent Detailed proofs of these results are given in \cite{ma}, in the following we outline the proof ideas of both theorems.
Similarly to Proposition \ref{prop_gen}, the following generalization holds.

\begin{proposition}
Let $k\ge 2$, $1\le \ell \le k-1$ and $n\in\N$ with $\tfrac{n-\ell}{k-\ell}\in\N$.
Let $\PP$ be a $k$-uniform $\ell$-path on $n$ vertices and let $c\ge 0$ be the smallest non-negative integer with $\frac{k+c}{k-\ell}\in\N$.
Furthermore, let $\alpha\in\R$, $1\le \alpha \le k$ be a constant and $W\subseteq V(\PP)$ be a vertex set such that for each edge $e\in E(\PP)$ we have $|e\cap W|\ge \alpha$.
Then
$$|W|\ge \frac{\alpha(n-\ell)}{k+c}.$$
\end{proposition}
\qed

A lower bound for $\ell$-paths with $\ell>\frac{2}{3}k$, Theorem \ref{lb_gen_large}, can be proven using an iteration similarly to Theorem \ref{lb_tight} with some minor modifications.
Since the above proposition is weaker than the bound obtained in Proposition \ref{prop_gen}, it is required in each iterative step
to divide the previous chosen $\ell$-path $\PP_{i-1}$ not into two parts $Z^1_i, Z^2_i$ on roughly $\frac{n}{2}$ vertices,
but into three parts $Z^1_i, Z^2_i,Z^3_i$ each of which has a size of about $\frac{n}{3}$. The remaining arguments are executed analogously to the proof stated above.
\\

For Theorem \ref{lb_gen_small}, when $\ell\le\frac{2}{3}k$, the proof of Theorem \ref{lb_tight2} can be applied.
Like in the mentioned proof we construct three edge-disjoint $\ell$-paths of different lengths. As sizes for the $\ell$-paths we choose $n_0=\left\lceil\frac{\ell(n-\ell)}{k+c}\right\rceil-1$, $n_1=n-1$, $n_2=n$.
Additionally, as a coloring for finding $\PP_1$ we assign the color red to the neighborhood $N_{>\ell-1}(E(\PP_0))$ (instead of $N_{>1}(E(\PP_0))$ in the above proof).
By the same arguments as before, we obtain the desired bound.

\section*{Acknowledgments}
The author would like to thank Mathias Schacht for his helpful comments and support with the thesis on which this paper is based as well as Maria Axenovich for comments on the manuscript.

redpajama_set_name: RedPajamaArXiv


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Hoping to land on The Ellen DeGeneres Show, SM East grad and friends document cross-country limo trip
Leah Wankum Aug 21, 2018 - Prairie Village
The team creating Ellen We Bought a Limo, from left, is David Sosna, Ty Gatewood, Jake Triplett and Kyle Brown.
Local videographer David Sosna spent part of his summer traveling across the country in a beat-up limousine. He and his friends are hoping to get on The Ellen DeGeneres show, but more importantly, they're giving back to the people they meet along the way.
The 2013 Shawnee Mission East graduate is a full-time freelance videographer, producing for commercial work and events. Mutual friends introduced Sosna to Jake Triplett, a YouTuber who used to work for K-Life in Prairie Village. During their introductory phone call at the end of June to discuss potentially working together, Triplett invited Sosna to join him and his friends on the cross-country road trip.
"It was bold on their side, but on my side, I think it would be crazy if I said no, and it would be crazy if I said yes," Sosna said of the offer. "I try to live on the 'why not?' of things."
About a week later, Sosna flew to Portland, Maine, to meet the other travelers and begin their journey.
The travelers began their cross-country trip in early July in Portland, Maine. Their journey ends in a few weeks in Los Angeles.
The idea behind their cross-country project was to drive through all 48 states in a beat-up 1990s limousine, with the end goal of getting on The Ellen DeGeneres Show. Their journey is slated to end in a few weeks in Los Angeles.
Sosna helped document the travelers' progress for a portion of the trip. He jumped off the trip a few weeks later in North Carolina because he received several gigs for his business in the Kansas City area.
The travelers just made their stop in Kansas City about a week ago. Sosna is considering re-joining the group when they reach Seattle.
The travelers' journey has garnered thousands of followers on their Facebook watch channel and Instagram page, "Ellen, We Bought a Limo." Some of their videos have upwards of 400,000 views; one popular video shows how they turned their limo into New York's hottest nightclub and drove around Times Square.
Sosna is sure DeGeneres has heard of them because they've received emails from her team. He's hopeful that by journey's end, they will receive an invitation to be on her show.
David Sosna, left, and Kyle Brown, another team member on the journey.
While getting on the Ellen show would be cool, the group's main goal is to give back to people in need whom they meet along their way. That's why they set up a GoFundMe for viewers to donate; 100 percent of those funds are distributed to people "who needed help in some way."
The first of these was a mother of two and McDonald's employee who needed financial help earning her GED diploma. Another was deep in medical bills for cancer treatment. Still another needed a leg up on her nonprofit that helps children meet with their parents who are incarcerated.
"We wanted to bring positivity everywhere that we went and realized that no matter where we went in the U.S., there are amazing, incredible people," Sosna said.


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The personal approach of Bonas MacFarlane and our particular philosophy takes our service above and beyond that of our competitors. We aim to help children succeed by providing a learning environment that is effective, inspiring and suited to their individual needs. Our priority is to instil children with the confidence to become independent learners. Many of the services we offer today, we pioneered and we have been leading the way ever since. Our consultants and tutors both in the UK and UAE have gained a reputation with discerning families for helping their children succeed at the leading schools and universities in Britain and the USA. Our approach is broad ranging, highly individual and discreet. We work with children in their family environments and also work closely with many schools.


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Home > Reviews	> ANNIHILATION – Ben Salisbury and Geoff Barrow
ANNIHILATION – Ben Salisbury and Geoff Barrow
February 28, 2018 Leave a comment Go to comments
Annihilation is an ambitious and intelligent sci-fi action-drama based on the series of Southern Reach novels by author Jeff Vander Meer. The film stars Natalie Portman as Lena, a biologist and former soldier, who is recruited by a secret US military science organization, and becomes part of a team sent to investigate 'the shimmer,' a mysterious phenomenon that has appeared in the Florida Everglades. Once inside the quarantined shimmer zone, Lena and her all-female team – psychologist Ventress, medic Anya, physicist Josie, and geologist Sheppard – begin to search for the source of the phenomenon, and encounter all manner of unusual animals and plants along the way. However, unknown to the other members of the group, Lena also has a secondary mission of her own: to find out what happened to her husband Kane, who was part of a previous team sent into the Shimmer, and which never returned. The film co-stars Jennifer Jason Leigh, Gina Rodriguez, Tessa Thompson, Tuva Novotny, and Oscar Isaac, and was directed by Alex Garland.
The film has been roundly praised for its innovation and its visual magnificence, as well as for its fascinating exploration of themes involving identity, the nature of reality, loss and grief, and for addressing complicated scientific topics including evolutionary biology and cellular chemistry. For those who don't care to get into all that, it's also just a really good yarn, with moments of action, tension and horror that work very well. I don't want to get too much into the detail, so as not to ruin the mystery, but the finale of the film will likely be divisive. I personally loved it, and enjoyed both the ambiguity and downright weirdness of it all. Unfortunately, the 'visual magnificence' element of the film has been somewhat undermined by film studio Paramount's decision to release the film straight-to-Netflix everywhere except the United States, a short sighted choice which will rob cinema goers of some of the most startling, memorable big screen imagery in quite some time.
The final aspects of the film which have been almost unanimously praised by mainstream critics are its sound design, and its score. The sound design elements, and the associated sound effects editing, are genuinely excellent, creative, and at times overwhelmingly powerful. As far as the score is concerned… well, let's just say I have a different opinion. The music is by British composers Ben Salisbury and Geoff Barrow. 46-year old Barrow is one of the founder members of the 'trip hop' electronica band Portishead, while 47-year old Salisbury is a composer who has worked extensively on numerous BBC nature documentaries. The pair teamed up for the first time in 2014 to write the score for Garland's directorial debut, Ex Machina, and collaborated again last year on the action thriller Free Fire.
Whenever I sit down to write a film score review, one of my first thoughts is to try to be fair to the composers. No score is written in isolation; it's informed by the narrative, the visuals, the wishes of the director (and the producer, and the executive producer…), and of course it must serve the story, first and foremost. When it comes to predominantly electronic scores like Annihilation I have to work extra hard on this, because my natural musical taste is geared towards acoustic instruments and orchestras, thematic ideas, melody, and so on. Obviously not every film can be scored like that, and many of them should absolutely not be scored like that. There are many examples where electronic sounds are required, where processed synthesizers should be used primarily, and where a themes-and-variations structure would not be appropriate. In those circumstances, I try to understand what it is the composer is actually saying with the music. What does it mean? What is this music doing to serve the narrative?
On something like Jóhann Jóhannsson's Arrival, which Annihilation superficially resembles, it was clear that Jóhannsson was writing music that acted similarly to the alien language in that film, and that he was using compositional palindromes as a way to musically mimic one of the story's key intellectual ideas. As such, as I was listening to Annihilation, and as I was watching the film, I was trying to figure out what Barrow and Salisbury were doing. Were they trying to address the intellectual concept of refraction that continually occurs in the film? Were they using some sort of music-based prism to somehow scramble and then re-assemble the DNA of their music, the way the Shimmer does to light, and time, and to the DNA of living organisms? And then something occurred to me… and that was that I'm working way too hard on this. If I'm having to look so closely, and dissect the score so minutely for these incredibly deep philosophical and technical connections to justify why the music in Annihilation sounds the way it does, then it's gotten way past the point where the music has failed to achieve the thing that in my opinion all film music must achieve, and that's support the film.
The bottom line is that, irrespective of all the intellectualizing and theorizing one could do, Annihilation is a failure of basic musical storytelling. I've written variations of this paragraph before but it bears repeating again: for me, the best film composers are great storytellers in the same way that the best directors are great storytellers. They take you on a cinematic journey, make you feel a myriad of emotions, they connect dots and narrative strands with recurring thematic ideas – be they melodic, textural, or whatever – and help guide the audience toward the experience the filmmakers want them to have. Annihilation, for me, does none of that.
Virtually the entire score – and I do mean the entire score – is a series of slow, droning tones augmented with bass thumps, synthesized ambient noises, and occasional little tinkles from various metallic percussion items. There are a decent amount of live, acoustic instruments buried in there somewhere – I detected what might have been some low brass in "Approaching the Shimmer," in addition to the beds of strings mentioned below – but whatever human touch is present has been digitally manipulated within an inch of its life so that all semblance of the original sound has been virtually obliterated. I don't possess the vocabulary to adequately describe what it really sounds like because, frankly, 90% of it sounds the same. They use different synthesizers to create different tones, but by the middle of the score it has basically all morphed into one all-encompassing drone. In fact, a lot of the score seems to exist for no other reason than the fact that it's necessary for there to be some sound, as opposed to there being silence, and they are using the ambiences to fill that void.
One brief melodic idea exists: a simple acoustic guitar riff with a slight country/bluegrass twang, which appears in "What Do You Know," "Disoriented," "The Watchtower," "Sheppard," "Cells Divide," and others. While watching the film I initially identified this as a recurring idea to represent the relationship between Lena and Kane, but then it appears in a couple of scenes in the film which have no clear connection to the Lena/Kane relationship, which means it's either not a theme for them, or that the director decided to use it in other random scenes without any regard for the composers' intended thematic application.
There are also a couple of notable textures which stand out. There is a humming vocal texture in "Ambulance Chase," "For Those That Follow," "Sheppard," "The Body," "Lighthouse Chamber," and others, although in the former it is immediately followed by a screaming, irritating electronic sound effect that made me want to rip off my headphones. Elsewhere, in cues like "The Swimming Pool" and "The Beach," Salisbury and Barrow make use of a bank of whining, keening strings, again augmented and overwhelmed by electronic textures, to create an atmosphere of overwhelming dread. I have read articles comparing the string writing in this score to the work of Krzysztof Penderecki but, truthfully, it's nowhere near as challenging, innovative, or compositionally interesting as the Polish composer's groundbreaking work – instead, it comes across as a pale, noisy imitation.
"The Bear" is unsettling, especially when it introduces a series of unearthly electronic groans and whines in its second half, and the first part of the aforementioned "The Beach" has a glassy sound that conveys a sense of crystalline wonderment and cool detachment, but the score's most significant cue is the 12-minute finale, "The Alien." It is in this cue that Salisbury and Barrow reveal what will likely be the score's defining musical idea – the three-note electronic pulse that featured heavily in the film's trailer, and which emerges for the first time at 4:47. They surround it with the most ambitious music in the entire score: iridescent strings, tortured-sounding brass, and bleating electronic tones which fade in and out and echo through our brains, as well as a reprise of the humming and sighing vocal texture, which at times appears almost angelic in tone. Some of the music is deeply, deeply unpleasant – there are long sequences where the music is little more than loosely organized static – but some of the more avant-garde electronic ideas are really quite impressive, especially when the music appears to be mimicking an otherworldly approximation of speech. If only the rest of the score had shown this level of innovation and creativity.
As you can clearly tell, with the exception of the 12-minute finale cue, I think Annihilation fails on all other fronts. It is neither a good film score in the narrative context of the film, nor is it an interesting piece of music in actual compositional terms. Salisbury and Barrow have their sound, and their way of working, and there is absolutely a place for this sort of writing in film music, as scores like Arrival attest. However, what bothers me most is that it's becoming predictable. Truthfully I'm getting sick of the prevailing wisdom in Hollywood that, in order for it to be taken seriously and project a veneer of intellectualism, serious sci-fi film music needs to have all its emotion, all its melody, and all its narrative structure, stripped away, so that all we are left with is are quiet ambient tones enlivened only by occasional moments of ear-drum shredding noise.
For me, serious science fiction is often about finding the humanity among the cool intellect, searching for emotional meaning in a galaxy that is often devoid of it. If that emotional context is removed from film music, if there is no depth of feeling bridging the gap between the film and its audience, then the music becomes nothing but an intellectual exercise. Then, when something like Annihilation comes along, which for me failed to be musically stimulating from either an emotional or an intellectual point of view, all you're left with is an hour's worth of hollow noise with no narrative arc, no emotional core to connect with, and only a single 12-minute finale cue that teases us with the sort of intellectual composition the whole score should have had.
Buy the Annihilation soundtrack from the Movie Music UK Store
What Do You Know? (2:36)
Ambulance Chase (2:58)
Approaching the Shimmer (1:53)
Disoriented (2:37)
The Alligator (1:02)
For Those That Follow (2:49)
The Swimming Pool (2:57)
The Watchtower (2:08)
Sheppard (2:45)
The Body (2:04)
Plant People (2:45)
Cells Divide (1:37)
The Bear (4:53)
The Beach (4:24)
Were You Me? (3:03)
Lighthouse Chamber (2:05)
The Alien (12:04)
Annihilation (5:22)
Lakeshore Records (2018)
Music composed by Ben Salisbury and Geoff Barrow. Recorded and mixed by Rupert Coulson. Edited by Yann McCollough. Album produced by Ben Salisbury and Geoff Barrow.
Categories: Reviews Tags: Annihilation, Ben Salisbury, Film Score, Geoff Barrow, Reviews
Lasse Vogt
Oh boy. Writing reviews like this is never easy. And I expect a lot of people to yell at you for having a well-rounded opinion, which is always annoying as hell. If that happens, you have my pitty. I am on your side.
Scott W Weber
I have to agree…as much as I like the movie, the score was a mess and I can't imagine listening to it outside of the film either.
EMPIRE OF THE SUN – John Williams PLANET OF THE APES – Jerry Goldsmith


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Mniszki () ist ein kleiner Ort im Nordosten der polnischen Woiwodschaft Ermland-Masuren. Er gehört zur Gmina (Landgemeinde) Budry (Buddern) und liegt im Powiat Węgorzewski (Kreis Angerburg).

Geographie
Mniszki liegt etwa 18 Kilometer nördlich der Kreisstadt Węgorzewo (Angerburg) unmittelbar im Grenzgebiet zu Russlands Oblast Kaliningrad (Gebiet Königsberg (Preußen)) und ist auf nur recht unwegsamer Straße von Ołownik (Launingken, 1938–1945 Sanden) aus zu erreichen. Der drei Kilometer entfernte Nachbarort war vor 1945 zugleich Bahnstation an der Strecke von den heute auf russischem Gebiet liegenden Städten Gumbinnen (russisch: Gussew) und Darkehmen (1938–1946 Angerapp, russisch: Osjorsk) bis nach Angerburg, die heute nicht mehr in Betrieb ist.

Geschichte
Das ehedem "Nonnenberg" genannte Dorf war vor 1945 wie auch Alt Eszergallen (1936–1938 Alt Eschergallen, 1938–1946 Sandenwalde), Neusorge und Wald Illmen (die beiden letzteren, nicht mehr existenten Orte lägen heute auf russischem Staatsgebiet) ein Ortsteil der Gemeinde Launingken, die zwischen 1874 und 1945 zugleich namensgebender Ort und Verwaltungssitz des Amtsbezirks Launingken war. Er gehörte – von 1939 bis 1945 in "Amtsbezirk Sanden" umbenannt – bis 1945 zum Landkreis Darkehmen (1939–1945 Landkreis Angerapp) im Regierungsbezirk Gumbinnen der preußischen Provinz Ostpreußen.

Im Jahr 1905 zählte Nonnenberg 33 Einwohner in nur einer Wohnstätte.

Infolge des Zweiten Weltkrieges kam Nonnenberg zu Polen, wurde 1945 in Mniszki umbenannt und wechselte vom Landkreis Darkehmen in den Powiat Węgorzewski (Kreis Angerburg), nun in der Woiwodschaft Ermland-Masuren (1975 bis 1998 Woiwodschaft Suwałki) gelegen. Mniszki gehört zur Landgemeinde Budry (Buddern) und ist dem Schulzenamt Olszewo Węgorzewskie (Olschöwen, 1938 bis 1945 Kanitz) zugeordnet.

Religionen
Bis 1945 war die überwiegend evangelische Bevölkerung von Nonnenberg in das Kirchspiel Dombrowken (1938–1945 Eibenburg, seit 1945: Dąbrówka) eingepfarrt. Es war Teil des Kirchenkreises Darkehmen (1938–1946 Angerapp, seit 1946: Osjorsk) in der Kirchenprovinz Ostpreußen der Kirche der Altpreußischen Union.

Die heute fast ausnahmslos katholische Einwohnerschaft von Mniszki gehört ebenfalls zur Kirche in Dąbrówka, die heute allerdings nicht mehr Pfarr-, sondern Filialkirche und dem Dekanat Węgorzewo (Angerburg) im Bistum Ełk (Lyck) der Katholischen Kirche in Polen unterstellt ist. Die evangelischen Kirchenglieder gehören jetzt zur Kirche in Węgorzewo im Pfarrsprengel Giżycko (Lötzen) innerhalb der Diözese Masuren der Evangelisch-Augsburgischen Kirche in Polen.

Einzelnachweise

Ort der Woiwodschaft Ermland-Masuren
Gmina Budry

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The youngest member of the Kardashian-Jenner empire has landed the cover of Forbes, and not for her good looks. The 20 year-old was named to Forbes' annual list of America's Richest Self-Made Women. The reality star fell into place number 27 and is darn proud of it. She has an estimated net worth of a whopping $900 million.
She has seen all of the success through her cosmetics line, Kylie Cosmetics. She is the sole owner of the massive company. She stated, "I'm ready to put up my own money. I don't want to do it with anyone else." Furthermore, Jenner is the youngest person ever to make it on the Forbes' list.
So congrats to Kylie Jenner who is showing some major woman power!


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PS5Xbox Series
Control: Ultimate Edition for PS5 and Xbox Series launches digitally on February 2, 2021, physically on March 2
Teaser trailer highlights Performance and Graphics modes.
Sal RomanoDec 18, 2020 at 9:08 AM EST0 Comment0
Control: Ultimate Edition will launch digitally for PlayStation 5 and Xbox Series on February 2, 2021, and physically on March 2, publisher 505 Games and developer Remedy Entertainment announced.
A new teaser trailer, recorded on PlayStation 5, highlights the new version's 60 frames per second "Performance Mode" and 30 frames per second with ray-tracing "Graphics Mode."
Control first launched for PlayStation 4, Xbox One, and PC via the Epic Games Store in August 2019, followed by PC via Steam in August 2020. Cloud-based versions of Control launched for Switch and Amazon's Luna service in October 2020.
Watch the teaser trailer below.
Fantasy puzzle adventure game Scarlet Hood and the Wicked Wood for PC launches in February 2021 Previous post
Say No! More delayed to spring 2021 Next post


Menyw is a variation of previous benyw, word related to Welsh banon 'maiden queen', Old Breton ban-doiuis 'goddess', Old Irish ban-chú 'female dog'; akin to Cornish benow 'feminine, female' and further related to Old Welsh ben 'woman', from Proto-Celtic *benā, from Proto-Indo-European *gʷḗn ("woman" [English queen, Norwegian kvinne, "woman", and kone, "wife", Greek γυναίκα (gynaíka, "woman; wife", thence gynecologist), Russian жена́ (žená, "wife"), Armenian կին (kin, "woman; wife"), Persian زن (zan, "woman; wife"), Sanskrit जनि (jáni, "woman") and ग्ना (gnā́, "wife") and Tocharian B śana, "woman, wife"]). Doublet of banw 'female'.
I've never heard menyw before so I'm assuming it's Southern? Aslo, I'm aware of yw but I feel like that's a Southern construction, am I right?
Well, I'll defer to you since I don't speak Welsh every day but I've only ever seen yw (unless I'm missing something it's normally ydy here) in writing here, up North and I've never heard anyone say or write menyw in about twenty years until I started this course.
Certainly seen both used on the BBC News/Cymru site recently. The media seem to use the various forms pretty flexibly.
I'm not sure how that backs up that it's 'widely used' if you're saying they use the various forms (including North/South-specific) flexibly.
Have to fight the urge to not write "she's a menu."
The only Welsh word for woman I'd encountered before was gwraig so now I'm thoroughly confused.
Any Welsh dictionary will usually show at least three words for 'woman'.


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Q: A disabled Input submit button color shouldn't change when the mouse is on it I have 2 buttons on a page and enable/disable of these buttons are done by evaluating a checkbox selection.
Currently the disabled button also changing the color when mouse is on them.
I want to control the .css button hover based on whether a button is disabled or enabled.
Do you know how to ensure that the colors shouldn't change for disabled button when the mouse is on them?
.cshtml
<input class="submit-btn submit-right-margin" id="case-escalate" type="submit" value="Update & Escalate" data-caseId="@Model.CaseId"/>

<input class="submit-btn submit-right-margin" id="case-update" type="submit" value="Update" data-caseId="@Model.CaseId"/>

.css
.submit-btn:hover
{
  color: #2474AD;
  background: #BFDDFF;
}

.js
I manage (enable/disable) input submit buttons based on a checkbox selection from a .js file.

A: You can use this to format disabled inputs:
input[disabled] {
    background:gray;
    color:silver;
    ...
}

and with input[disabled]:hover you can format the hover state.
But the [disabled] is still not working on a few browsers.
EDIT:
Never tested this before:
Maybe you can handle this with classes. Just add/remove a class when disabled/enabled with JavaScript.

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Nebraska Law Review
Home > Law > NLR > Vol. 84 (2005) > Iss. 1
Kennewick Man, Kinship, and the "Dying Race": The Ninth Circuit's Assimilationist Assault on the Native American Graves Protection and Repatriation Act
Allison M. Dussias, New England School of LawFollow
This article examines the Bonnichsen case and the Ninth Circuit's controversial and fundamentally flawed rewriting of the Native American Graves Protection and Repatriation Act (NAGPRA), in light of the history of Native Americans' encounters with anthropologists and of the goals that NAGPRA was intended to achieve. Part II discusses the history of anthropologists' activities involving Native Americans and the remains of their deceased kin, revealing the longstanding practices that NAGPRA was enacted to redress. This historical background provides the context in which NAGPRA must be understood and interpreted, and in light of which decisions under the statute should be made. Moreover, while contemporary anthropologists may claim that this history should be forgotten because it reflects attitudes that they believe their profession has repudiated, for tribal members who continue to be subjected to anthropologists' claims to the remains of their ancestors, this past is still very much a part of the present. Part III examines NAGPRA itself and the key determinations that need to be made when the statute is applied to human remains. Part III also discusses the DOI's application of the statute to the Kennewick remains. Part IV examines the views of the plaintiffs and the defendants in the Bonnichsen case and critiques the Ninth Circuit's 2004 decision in favor of the plaintiffs. A careful analysis of the court's language reveals much about the underlying attitudes and sympathies that shaped the outcome. This analysis also demonstrates how the court's approach to the case and the case's outcome both reflect continuing rejection and subordination of the Native American perspective that Congress, in enacting NAGPRA, established as being entitled to respect, and thus perpetuate the assimilationist attitude that for so long characterized government policy toward Native Americans. The Conclusion offers some final thoughts on the Bonnichsen decision and the future treatment of Native American remains, and on the continuing survival of the "dying race," against seemingly great odds.
Allison M. Dussias, Kennewick Man, Kinship, and the "Dying Race": The Ninth Circuit's Assimilationist Assault on the Native American Graves Protection and Repatriation Act, 84 Neb. L. Rev. (2005)
Available at: https://digitalcommons.unl.edu/nlr/vol84/iss1/3
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Nebraska Law Review Bulletin Archive


text:
import argparse
import fnmatch
import os
import sys
import time
import xml.etree.ElementTree as ET

def find_files(path, pattern):
    matches = []
    for root, dirnames, filenames in os.walk(path):
        for filename in fnmatch.filter(filenames, pattern):
            matches.append(os.path.join(root, filename))
    return matches

if __name__ == "__main__":

    parser = argparse.ArgumentParser()
    parser.add_argument("jobs", nargs='+', help="jenkins jobs containing nosetests.xml somewhere")
    parser.add_argument("--name", "-name", default="Consolidated Test Report", help="report name at beginning of email subject line")
    args = parser.parse_args()

    JENKINS_HOME = os.getenv('JENKINS_HOME')
    if JENKINS_HOME is None:
        print "JENKINS_HOME not set.  Exiting."
        sys.exit()

    results = {}
    total_tests = 0
    total_failures = 0
    total_errors = 0
    total_skip = 0
    total_pass = 0
    # set() to remove duplicate jobs
    for job in set(args.jobs):
        nose_reports = find_files(os.path.join(JENKINS_HOME, 'jobs', job), 'nosetests.xml')
        if nose_reports:
            # using first nosetests.xml found
            tree = ET.parse(nose_reports[0])
        else:
            continue
        test_suite = tree.getroot()
        results[job] = test_suite.attrib
        results[job]['tests'] = int(results[job]['tests'])
        results[job]['failures'] = int(results[job]['failures'])
        results[job]['errors'] = int(results[job]['errors'])
        results[job]['skip'] = int(results[job]['skip'])
        results[job]['pass'] = results[job]['tests'] - results[job]['failures'] - results[job]['errors'] - results[job]['skip']
        results[job]['success_rate'] = float(results[job]['pass']) / (results[job]['tests'] - results[job]['skip'])
        results[job]['date'] = time.ctime(os.path.getmtime(nose_reports[0]))

        total_tests += results[job]['tests']
        total_failures += results[job]['failures']
        total_errors += results[job]['errors']
        total_skip += results[job]['skip']
        total_pass += results[job]['pass']
    try:
        total_success = float(total_pass) / (total_tests - total_skip)
    except ZeroDivisionError:
        # no results or all tests skipped
        total_success = 0.

    # write file with summary line for email
    with open('email_subject.txt', 'w') as f:
        summary_line = args.name + " Tests: " + str(total_tests) + " Failures: " + str(total_failures) +  " Errors: " + str(total_errors) + " Success Rate: " + '{:.2%}'.format(total_success)
        f.write(summary_line+'\n')

    # write file for email body
    with open('email_body.html', 'w') as f:
        f.write('<!DOCTYPE html>\n')
        f.write('\n')
        f.write('  \n')
        f.write('    <meta http-equiv="Content-Type" content="text/html; charset=utf-8">\n')
        f.write("""    """)
        f.write('\n    <title>Test Results</title>\n')
        f.write('  \n')
        f.write('  \n')

        f.write('
Summary
\n')
        f.write('
\n')
        f.write('
\n')
        f.write('        Tests	\n')
        f.write('        Failures	\n')
        f.write('        Errors	\n')
        f.write('        Success Rate	\n')
        f.write('
\n')
        f.write('
\n')
        f.write('        ' + str(total_tests) + '	\n')
        f.write('        ' + str(total_failures) + '	\n')
        f.write('        ' + str(total_errors) + '	\n')
        f.write('        ' + '{:.2%}'.format(total_success) + '	\n')
        f.write('
\n')
        f.write('
\n')

        f.write('
All Tests
\n')
        f.write('
\n')
        f.write('
\n')
        f.write('        Job	\n')
        f.write('        Tests	\n')
        f.write('        Failures	\n')
        f.write('        Errors	\n')
        f.write('        Success Rate	\n')
        f.write('        Date	\n')
        f.write('
\n')
        for job, job_results in sorted(results.iteritems()):
            f.write('
\n')
            f.write('        ' + job + '	\n')
            f.write('        ' + str(job_results['tests']) + '	\n')
            f.write('        ' + str(job_results['failures']) + '	\n')
            f.write('        ' + str(job_results['errors']) + '	\n')
            f.write('        ' + '{:.2%}'.format(job_results['success_rate']) + '	\n')
            f.write('        ' + job_results['date'] + '	\n')
            f.write('
\n')

        f.write('
\n')
        f.write('  \n')
        f.write('\n')

redpajama_set_name: RedPajamaGithub


An FA statement read: "It is alleged that his language at the end of the fixture, as captured by the broadcast footage, was abusive and/or insulting and/or improper.
"He has until 6pm on October 19 to respond to the charge."
The match was played amid a tense atmosphere with the club having denied a report earlier in the day that Mourinho was going to be sacked that weekend regardless of the result.
United recovered from 2-0 down to complete a dramatic comeback with a 90th-minute goal from Alexis Sanchez.
The 55-year-old, who claimed after the match that he was the subject of a media "manhunt", said when asked about the gesture: "It is a finger.
"Smaller than the others, but it is a finger."


text:
Defence Secretary announces fund for battlefield apps
The Defence Secretary has announced a new fund to develop battlefield apps able to respond quickly to high-pressure scenarios.
The £30million fund will go into developing battlefield apps and software for the Armed Forces capable of being adapted to the specific demands of missions and commanding officers and being put on a central network in a matter of days. The money is being drawn from the Transformation Fund and will form part of the Modernising Defence programme.
The Platform for Rapid Exploitation of Digital Platforms (PREDA) is similar to technology already used by the US Air Force and US Marine Corps, which has already produced positive results. The adaptability of the technology means that it can be changed to accommodate new threats and changing conditions quickly, making it ideal for extreme environments like disaster response.
Defence Secretary Gavin Williamson said: "From major natural disasters to the frontline on the battlefield, our Armed Forces are quick to react and even quicker to adapt. This technology matches that agility by enabling the creation of specialist digital applications that are tailored, secure and responsive."
"By deploying this technology into the hands of our military, from sailors serving in the Caribbean, to pilots working in the Middle East, we will ensure we have a fighting force fit for the future, equipped with state-of-the-art technology at their fingertips."
The technology is expected to be in use by the end of this year. The announcement follows a £66million fund for military robotics promised by the Defence Secretary last week.
The project is backed by Joint Forces Command and is expected to ultimately be rolled out across all departments of the armed forces. It will initially be used in decision making and command.
The post Defence Secretary announces fund for battlefield apps appeared first on Defence Online.
New Gurkha battalion to begin recruiting this year
"Cyber security threat goes far beyond Russia" according to the NCSC


text:
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The micemd package. It's crantastic!
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text:
HomeArtsWONY DJ Spotlight: Alan Stevens
WONY DJ Spotlight: Alan Stevens
April 12, 2019 The State Times Arts 1
Talia Bautista, Contributing Writer |
Alan Stevens is a junior at SUNY Oneonta majoring in mass communications. He joined WONY in the fall semester of 2018 and began hosting "King Baby Alan: The Show" last November.
When Stevens learned about WONY as a freshman, he at first was intrigued to join but didn't because it seemed daunting. As he became friends with more people at SUNY Oneonta, he ended up becoming a figure in the music scene by going to house shows and befriending artists. Soon enough, Stevens met people who were part of WONY, and they wanted him to join, thinking he would have a good show. Stevens even sat through a couple of shows on the station as a guest before joining.
Last semester Stevens finally decided to give WONY training a shot to see where it could go. He notes that it was intimidating, but he ended up completing his training and passing the tests. He immediately began to plan out his own show. At first, Stevens didn't know what to call it. Alec Feldherr nicknamed him "King Baby" as a unique moniker within their friend group. The show name formed as a throwaway title while bouncing ideas around with another friend, Jonah. He was the one who convinced Stevens to use that name.
Music was one of the only things Stevens said he ever really cared about throughout middle school and high school. His show's playlists revolve around what he's listening to throughout the week. One show could be about punk, while another could consist of electronic music. At first, Stevens was playing a ton of different genres, doing the sonic equivalent of throwing everything at a wall to see what sticks; but now he usually focuses on one genre per show. Stevens also plays Top 200 music, as well as local acts.
Stevens notes, "Joining WONY was one of the best decisions I've made at Oneonta. The club is filled with so many wonderful people. It's also the only club that took me in with gigantic, open arms. The amount of work and dedication you put into it pays off tremendously as well."
Listen to "King Baby Alan: The Show" every Tuesday from 6 p.m. to 8 p.m. on wonyfm.com or the tune in app!
Interested in joining WONY? Contact Angie, our Training Director at [email protected], and don't forget to Keep It Locked!
Review: WONY Island
Virginia Wins Program's First National Championship
WOW. I am stoked to read this. I remember King Baby Alan as the little man climbing ladders to get to the roof before his parents got to him (age 3?). The little man that would drop from the second floor to the first just to scare us again around 3 years old. Yup I am stoked to read this. What a great accomplishment in life. So happy for our little man, your King Baby. Def will listen to his work. Congrats Alan!!!!


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Eating more heart-healthy omega-3 fats provided no additional benefit in a study of heart attack survivors who were already getting good care, Dutch researchers report.
After nearly 3 1/2 years, there was no difference in deaths, heart attacks and other heart problems between those who ate margarine with added omega-3 fatty acids and those who didn't, the study found.
The results don't mean that getting more of the essential nutrient has no value. Several studies have offered evidence that the fats — mostly from fish oil — reduce heart disease.
"Now they're popping up in the most unexpected places," including trail mix, said Lichtenstein.
It's generally recommended that people eat one or two servings of fish a week, said the study's leader, Daan Kromhout of Wageningen University in the Netherlands.
"The results of this trial do not change that," he said, noting that there were no harmful side effects. "It's still a good thing to eat fish once or twice a week."
The study's participants were taking the best medicines to prevent future heart trouble, and that could be why adding a low-dose of omega-3s offered no extra protection, the researchers said. The volunteers were also older and entered the study years after their heart attacks, in contrast to heart patients in earlier research who did benefit by taking fish oil pills.
For the study, Kromhout and his colleagues recruited 4,837 heart attack survivors in the Netherlands who were 60 to 80 years old. They were randomly assigned to eat one of four kinds of margarine — regular margarine or ones with added omega-3s derived from fish, plants or both.
Dr. Stephen Kopecky, a cardiologist at the Mayo Clinic in Rochester, Minn., said it may have been a matter of too little, too late — the dose was small and the patients were enrolled many years after their initial heart attack — on average four years.
Since the Dutch study was in heart attack survivors, Lichtenstein said it still isn't known whether omega-3s can protect against a first heart attack or help those patients who don't get such good care.


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text:
Credit: Dinner Series/flickr
Will There Ever Be a Five-Star Restaurant Review?
Besha Rodell February 18, 2014
There's a question I've been asked now by friends, co-workers, readers and even my editors: Are you ever going to give a five-star review?
It's a fair enough question. It's been just over a year since we started awarding star ratings on our restaurant reviews, and although we use a five-star scale, there have been no five-star reviews. This isn't a big change for me: In my last job, as a critic in Atlanta where I was also on a five-star system, I gave only one five-star review in six years.
So why even bother having five stars if we're never going to use them?
The answer: While we have a ton of restaurants here that we should be immensely proud of, Los Angeles should be held to a global standard, the same as New York or London or Paris. We deserve restaurants that travelers might plan a trip around, that might end up on someone's international bucket list of dining. While I've reviewed some incredibly impressive restaurants in the last year, none of them quite met that standard. Some came very close, and one in particular might turn into a five-star restaurant one day. But not yet. And that's OK.
In the past year, I've heard the argument that if we never use the fifth star, it loses all meaning. I tend to believe the opposite – that if we give out stars too liberally, they lose meaning. The New York Times currently has only six four-star restaurants in the entire city, and that's over years and years of reviews, and quite a few critics. It's part of what makes stars in the Times so meaningful – they are not given easily. When that five-star restaurant comes around, you'll know how special it is.
Think of the life-changing meals you've had, the ones you'll never forget, the ones that were perfect in every way from food to wine to service, or so unique they moved you in a way you never expected a meal could. How often in your life do you expect to have such a meal? I can count on one hand the meals I've had like that, and I eat out for a living. When something like that does come along in Los Angeles, shouldn't there be a way to distinguish it? A way to separate it from the merely wonderful?
The truth of the matter is that, until that world-class restaurant opens, I'm basically operating on a four-star scale. There is a range within each star rating, but as with a four-star scale, a three-star rating should be seen in many instances as a rave, and a four-star rating should be seen as an indication that this is one of the very best places in town. Two stars should be seen as an endorsement, and one star, while probably never positive, should not be seen as a wholly negative review – the rating means "fair," which means the restaurant is OK. Not good but not bad.
So, the short answer to the question is that I probably will give a five-star review one day. But not simply because we have the option – when the fifth star happens, it will be to honor something truly astounding.
Have a question for the critic? Email them to brodell@laweekly.com.
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text:
Maré steht für:

 eine Favela im urbanen Gürtel von Rio de Janeiro, siehe Maré (Rio de Janeiro)
 ein Musikprojekt für behinderte Laienmusiker, siehe Maré (Musikprojekt)
 einer Insel der Loyalitätsinseln, Neukaledonien, siehe Maré (Neukaledonien)
 Maré, eine deutsche Metal- und Hardcore-Band aus Lingen/Münster, siehe Kalypso (Musikgruppe)

Mare oder Maré ist der Familienname folgender Personen:

 Catya Maré, deutsche Komponistin, Musikproduzentin, klassische Crossover Geigerin, bildende Künstlerin und Schriftstellerin
 Johann Karl Mare (1773–1835), deutscher Kartenstecher und Verleger

Siehe auch:
 Mare
 Marées
 Marais (Begriffsklärung)


The 10 Best Avengers Infinity War Merchandise Items Available Now!
Today is the day – Avengers: Infinity War movie is out in theaters! When I was in Los Angeles with Disney and Marvel at the movie premiere, I got the chance to check out a lot of the fun merchandise products that they've made for the movie, they have a merchandise room at Marvel's headquarters in which they had Infinity War products on display. I was lucky enough to take some home, too! From Avengers t-shirts to toys, mugs, and everything in between, here are some of my favorite Avengers Infinity War merchandise items.
This electronic fist is super cool!Wear it like a glove and use Thanos' fist to rule the galaxy!
These water bottles are the perfect Avengers Infinity War merchandise products for fans! Carry your water in style wherever you go.
When you go see the movie, you'll see which superhero uses this axe as a weapon.
People LOVE these little Funko Pop figurines, they are really cute versions of all of the films characters!
Dress up your office chair with this fun Spider Man cape. Super powers not needed!
Show how much you love the Avengers with a t-shirt! There are lots of designs and sizes available for all ages, choose your favorite!
This InfinityWar merchandise set has 10 detailed character figures, including Iron Man, Thanos, Gamora, Spider-Man, Thir, Star-Lord, Groot, Doctor Strange, and more! It includes pretty much every superhero in the movie, so you can re-enact the whole Infinity War story!
I saw this mug in person at Marvel merch room and it is beautiful! It's a replica of Thanos' gauntlet, is full of marvelous details, and has all 6 Infinity stones in it. I would love drinking my coffee out of this powerful mug every morning.
Avengers: Infinity War is now in theaters everywhere!