text:
music UX
the user experience of digital music
The transactional flow of Sub Pop Records
Sub Pop is a legendary record label based in Seattle that has signed great bands like The Shins, The Postal Service and Nirvana. They recently celebrated their 2oth anniversary and they have a great site full of artist info, photography and a Listen feature which allows users to stream music. However, when purchasing music, the transactional flow of the site is confusing.
After adding an album (in MP3S format) by the Vaselines to the the shopping cart, I noticed several options including login, create an account, a delete icon, a link to the album and a checkout icon. However, a few of these items were unclear. I'm not sure why one would want to create an account (from the shopping cart) and there is no helpful information on the subsequent 'Sign Up' screen to explain the benefit or creating an account. The login and account options would be better placed outside the context of the shopping cart area or simply introduced at the beginning of the checkout process.
I finally clicked the Checkout icon and began the transaction process. But on the third screen, I was asked to enter a shipping address! I thought I was downloading the digital MP3S format so why would I need to provide a shipping address? The site should contextually recognize that I am purchasing a digital version of an album and bypass the shipping address screen. This type of confusing transactional flow will frustrate users and may cause them to abandon the shopping cart altogether.
Sub Pop should minimize the number of choices in the Account area by removing login and create an account and make the flow more contextual by requiring users to enter only the information they need for that purchase. These changes would improve the overall music user experience of the site.
(Sub Pop Records gave me permission to use the image.)
Google Doodle tribute to Les Paul
In honor of (what would have been) Les Paul's 96th birthday, today's Google Doodle is an interactive, recordable, electric guitar. Paul, after whom the Gibson Les Paul guitar is named, was one of the first electric guitars and he designed one of the first solid body electric models. He is also one of the few artists with a stand-along exhibit at the Rock and Roll Hall of Fame . The Google Doodle, found at google.com , is typically an illustration created for one day to (among other things) celebrate a holiday or the birthday of someone famous. You simply hover over the strings to strum and create a song. A record button lets you save your song and send it to someone and you can even play back your own song while playing over it (like the overdubbing Paul made famous in the '40s). Click on the Doodle? Well, that will show you search results for Les Paul. Several music fans have recorded their own songs today including this one - 'Hey Jude' by The Beatles. Easy to u
A site map for a music application
For the past several months, I've been working with the CHIRP Radio volunteer tech team to design a mobile solution that allows station DJs to plan out their weekly shows. At a high level, a DJ should be able to browse and select songs from a massive music library, and add them to one or more playlists, which can be used for their shows. The app also has functionality like the ability to read album reviews, view recent activity by DJs in the app, and review a DJ's profile. A site map is an important artifact when designing any application or website. It shows how the overall navigation should be structured, can be used with end users to validate the taxonomy, and is helpful for developers as a companion to wireframes or mockups. Below is a site map I recently created for the project. The highest level navigation options are lighter in color, while as the user navigates deeper, darker colors are used to represent those options. The coloring isn't necessarily a known
Setting a price alert on Stubhub
A few weeks ago, my wife and I really wanted to see Alvvays and Frankie Rose at Metro in Chicago. But the show sold out and I was left to look for tickets second-hand. I've used Stubhub before to purchase concert tickets but up until now, hadn't tried their Price Alert feature which lets you set a ticket price max and be notified when the price goes below it. The user experience from beginning to end was really good. The feature is easy to find and provides very mobile-friendly controls to create the alert. The Price Alert tab was prominent on the event details screen - very easy to find. Note: today (1/5/18), Stubhub has removed those tabs and you have to tap an Info button top right of the viewport to look for the same feature. It's a still a good UX once you get there but it's an additional step. The Price Alert feature is intuitive and uses tappable numbers to let me pick the quantity of tickets (not a clunky dropdown menu or less efficient plus/minus p


A curtain is drawn open. Inside an intimate view of beauty. Grace.
Exhibited at The PaperGraphica Gallery Christchurch, with much aclaim.
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text:
Putin's Favorite Neighbor Whips Out the Big Guns in New Warning
Shannon Vavra
ALEXEY DANICHEV/SPUTNIK/AFP via Getty
Belarus' army has begun removing armored vehicles from long-term storage to contribute to the joint regional grouping of Russian and Belarusian fighting forces, according to the Belarusian Ministry of Defense.
The decision was made in order to facilitate "interoperability measures" for the battlegroups, which have been conducting joint drills since October in Belarus, according to TASS.
It's just the latest sign that Belarus could be gearing up for military action—or that it might be preparing to serve as a launchpad for Russian assaults on Ukraine.
Last fall, Russia sent tens of thousands of troops and armored vehicles into Belarus to conduct live fire exercises. Since then, Belarus has initiated snap military drills and moved military gear and equipment toward its border with Ukraine. Earlier this month, the Commander-in-Chief of Russia's Ground Forces, Army General Oleg Salyukov, also visited Belarus to inspect the joint Russian-Belarusian military force.
Belarusian leader Alexander Lukashenko allowed Moscow to use Belarus as a launchpad for its failed offensive against Kyiv last year, and has likely continued his attempt to beef up Belarusian combat readiness due to pressure from Russia, analysts say.
Russia Sets Ultimatum to Formally Pull a Third Country Into Putin's War
The news that Belarus is transitioning military gear out of long-term storage comes just days after Russia's foreign ministry outlined conditions under which Belarus would join the war in Ukraine. If Belarus or Russia believe that Ukraine has "used force" against either country, Belarus will join the war, according to the ministry.
It's not clear what "use of force" comprises in the eyes of Minsk or Moscow, but Lukashenko has been opining publicly about perceived Ukrainian threats to Belarus in recent days.
Lukashenko suggested Tuesday that Ukraine continues to make threatening moves by training armed forces to defend against a Belarusian assault on Ukraine from the north.
"On the one hand, they ask us not to send troops to Ukraine… On the other hand, they are preparing this explosive mixture and arming them," Lukashenko said Tuesday, according to Belta, adding that Kyiv proposed Belarus enter into a "non-aggression pact" with Ukraine. "Therefore, we are forced to react harshly… They should consider themselves warned."
It was not immediately clear if Ukraine had indeed suggested a "non-aggression pact" with Belarus. Ukraine's Foreign Ministry did not immediately return request for comment.
Ukraine's Foreign Ministry has sought to simmer tensions with Belarus, reassuring Minsk in a statement Tuesday that Ukraine does not intend to attack the country.
"We confirm the absence of any aggressive intentions towards Belarus on our part," the Foreign Ministry spokesperson, Oleh Nikolenko, told Ukrainska Pravda.
The Kremlin is likely working to pressure Belarus to prepare to engage in the conflict more, Ukraine's Foreign Ministry assesses.
"Russia is trying in every possible way to draw Belarus directly into a war. Ukraine warns Minsk against any possible further aggressive plans," Nikolenko said.
Ukrainian President Volodymyr Zelensky also issued a statement in an attempt to signal to Lukashenko that he ought to resist pressure to join the war.
"We were not going to and are not going to attack Belarus. This is the main signal from the entire Ukrainian people to the Belarusian people," Zelensky told reporters Tuesday. "It is very important for us that Belarus does not lose its independence and does not join this disgraceful war despite anyone's influence."
Qulliq Energy stops short of labelling cyberattack another Nunavut ransomware incident
The Qulliq Energy Corp. says it was locked out of its data in January's cyberattack, but stopped short of calling it a ransomware attack. QEC announced last month it fell victim to a cyberattack, which was discovered on Jan. 15. While no operating technology, such as power plant infrastructure, was affected, QEC's information technology, like email, billing and payroll databases were. "There's different levels of cyberattacks," said Bill Nippard, QEC's vice-president of operations and engineerin
DETROIT (Reuters) -General Motors Co shares jumped early on Tuesday after it reported higher net income for the fourth quarter, forecast stronger-than-expected earnings for 2023 and said it would cut $2 billion in costs. The automaker, the top in the U.S. by sales, forecast that it could hold its pre-tax margins steady between 8% and 10% through 2025, despite a price war that Tesla Inc has triggered in the electric vehicle segment. GM plans to build only about 400,000 electric vehicles in North America between now and the first half of 2024.
French finance minister backs pension push, UAE's COP28 pick
ABU DHABI, United Arab Emirates (AP) — Trying to sell France to Emirati investors Monday, France's finance minister highlighted what he described as the business-friendly overhaul of President Emmanuel Macron — while sidestepping ongoing protests back home over a planned pension overhaul. Finance Minister Bruno Le Maire's trip to Abu Dhabi, the United Arab Emirates' capital, comes as over 1 million people marched days earlier to protest a plan by Macron's government to increase the retirement ag
Former UK PM Boris Johnson claims Vladimir Putin said he could hit him with missile
In a new interview, former UK Prime Minister Boris Johnson said President Vladimir Putin did not seem serious about avoiding war in the days before Russia invaded Ukraine.View on euronews

redpajama_set_name: RedPajamaCommonCrawl


text:
Early ColourPop Employees Are Going Viral Again With Socially-Savvy Hair Brand 'Insert Name Here'
On May 16, 2019 May 16, 2019
A six-month-old hair brand called Insert Name Here — which vends wigs, extensions, and hair pieces on the web — has clinched a seven-figure annual run-rate after being founded in November. Social influencers, its three founders say, have been formative to the startup's success. And the three founders know a thing or two about social influencers.
Insert Name Here (INH) is looking to follow in the footsteps of socially savvy stalwarts like clothing retailer Fashion Nova and ColourPop Cosmetics, where two of INH's co-founders — Jordynn Wynn and Sharon Pak — were the first two employees.
Wynn and Pak helped ColourPop become a major player in today's beauty industry — thanks in no small part to its mastery of Instagram. Early on, for instance, ColourPop latched onto viral hashtags to amass hordes of followers, sponsored influencer posts, and eventually forged full-fledged product collabs with the likes of Bretman Rock, Amanda Steele, Zoe Sugg, Eva Gutowski, and Kathleen Lights.
Today, Wynn and Pak — alongside INH's third founder, Kevin Gould, a serial entrepreneur and digital vet in his own right — are hoping to trod a similar path in the hair category. Despite the fact that there has been some momentum in the space — twin YouTube stars Niki and Gabi DeMartino and trans beauty vlogger Nikita Dragun have all launched hair extension collabs with a competitor called Bellami — the trio still saw a vast opportunity.
"We thought it was an incredibly stagnant industry that lacked a strong digital presence and social community," Gould tells Tubefilter. While color cosmetics may be somewhat saturated, Gould says hair is still a wide open lane.
The company's name is a nod to the fact that hair accessories have a transformative ability to turn wearers into whoever they want to be. And thus far it's self-funded. Gould says that there are no current plans to take on outside funding, and the fact that INH is built on their own capital and time commitments forces them to be more disciplined.
While Wynn and Pak ascended at ColourPop, Gould got his start 10 years ago in the mailroom at talent agency WME. He then went on to become an early investor and founding partner at Beautycon – which hosts festivals for beauty influencers across the globe – before becoming an angel investor with a handful of other startups, including Jukin Media and Draft Kings. Today, he is the founder of Kombo Ventures — another digital media startup that operates separately from INH.
For its part, Kombo serves as a talent management company for YouTube stars like SSSniperWolf and Alex Lange; furnishes a digital strategy agency for Fortune 500 companies; and is also working to incubate its own influencer-led brands (the first of which, Gould says will debut later this summer with a big name in the digital space).
Gould isn't the only one moonlighting with another venture: Wynn is still employed by ColourPop, while Pak serves as the marketing director at another beauty brand called Il Makiage.
An Approachable Ethos
INH officially launched last November after the trio remarked that Gen Z and millennial shoppers were becoming increasingly interested in extensions and wigs, though prices for hair accessories were traditionally sky-high. INH's best seller, on the other hand, is the Miya Ponytail — an attachable piece priced at $45 that's made from Japanese fibers and can be heat-styled and washed (but not dyed). The company also vends 100% human hair extensions, ranging in price from $195 to $220 per set, depending on their length.
On top of an accessible pricing model, INH is also seeking to hone an approachable ethos by sharing hair tutorials and inspiration for those who are unfamiliar with fake hair use.
The company has nabbed 29,000 followers on Instagram, where its account shares user-generated tutorials and other eye-catching looks — many of which are sourced from the #INHBabe hashtag. INH also shares video tutorials on its website as well as other how-to content on a dedicated blog.
Kathleen Lights wearing the company's 'Zooey' bangs, $79.
During their time at ColourPop, Pak and Wynn built substantial Instagram followings on their personal accounts — to the tune of 74,300 and 73,300 followers, respectively. And in addition to using their channels to promote INH, they are also amplifying INH's own presence with their proclivity for snackable posts and keen ability to weed through the influencer space, carefully selecting partners that align with its values.
"We did not attach one big influencer to launch INH," Gould says. "Instead, we are leveraging support across the board."
To date, the brand's Instagram account has featured names like YouTube star Kathleen Lights (who counts 2.1 million followers on Instagram and is pictured above), model-entrepreneur Arianny Celeste (3.1 million followers), and Twitch star Hailie Barber (337,000 Instagram followers). The next step, Gould says, will be to team with a major influencer on an INH product collab — though the nascent company is still exploring options, with no names firmly in play.
At the end of the day, however, Gould says that having a masterful Instagram strategy isn't the only factor in launching a consumer brand, given the saturation of social platforms today.
"Instagram is still an incredibly important part of the template, but now, more than ever before, founders need to understand data on the back-end as well as all of the other elements of the marketing mix, including email acquisition, text messaging, and targeted advertising," Gould says. "It was only because of our collective backgrounds and experiences that we were able to launch and scale so quickly."
Bretman Rock Nods To Filipino Heritage With New ColourPop Collaboration
Zoe Sugg Collabs With ColourPop On Brunch-Themed Makeup Collection


text:
in every sense of the word.
With a roar like a train in a tunnel, the war enters its final days. Ellie has to be on that train. There's no more sitting around, no more waiting. There's only fast decisions, fast action, fast thinking.
And she'd better get it right. With all their being, Ellie and her friends want to stay alive. But there's a war out there. And it has to be won. The end is here...
Fans of Veronica Roth, Suzanne Collins and John Flanagan will love John Marsden.
John Marsden is Australia's bestselling author for teenagers and a highly acclaimed picture book writer. His previous titles include Winter, The Head Book, The Boy You Brought Home and Millie. John Marsden lives at the Tye Estate, just outside of Melbourne, where he opened a school called Candlebark in January 2006. The school currently has 52 students, ranging from Prep to Year 7.


Discover how your Percell ancestors earned their living and spread across the country. Find out more about your family history by using the Free Search box.
The surname Percell is the 4th most popular variant of Purcell.


text:
\section{Introduction} \label{sec1}
Let $R$ be a left and right noetherian ring (not necessarily
commutative), all modules left $R$--modules and $C$ a semidualizing
$(R, R)$--bimodule (Definition~\ref{d1}). A complex $X_\bullet$ of
$R$--modules is said to be $C$--{\it perfect} if it is
quasiisomorphic to a finite complex
$$T_\bullet= 0\longrightarrow C\otimes_R P_n\longrightarrow C\otimes_R P_{n-1}\longrightarrow\cdots\longrightarrow
C\otimes_R P_1\longrightarrow C\otimes_R P_0\longrightarrow 0,$$
where each $P_i$ is a finite (i.e. finitely generated) projective
$R$--module. The {\it width} of such a $C$--perfect complex
$X_\bullet$, denoted by $\mathrm{wd}(X_\bullet)$, is defined to be
the minimal length $n$ of a complex $T_\bullet$ satisfying the above
conditions. Recall from \cite{bf}, a ring $R$ is called {\it
strongly regular} whenever there exists a non-negative integer $r$
such that every $R$--perfect complex is quasiisomorphic to a direct
sum of $R$--perfect complexes of width $\leqslant r$. Buchweitz and
Flenner, in \cite{bf}, characterize the commutative noetherian rings
which are strongly regular.

Our first objective is to detect when a ring is strongly
regular by means of $C$--perfect complexes (Theorem~\ref{J}).
We also prove that $C$--projective modules (i.e. modules of the form
$C\otimes_R P$ with $P$ projective) have the ability to detect when
a ring is hereditary (Proposition \ref{H}).

Our second goal is to find a class of $R$--modules which admit minimal resolutions of
$C$--projective modules (see Theorem \ref{E}).

\section{Preliminaries} \label{sec2}

Throughout, $R$ is a left and right noetherian ring (not necessarily
commutative) and let all $R$--modules be left $R$--modules. Right
$R$--modules are identified with left modules over the opposite ring
$R^\mathrm{op}$. An $(R, R)$--{\it bimodule} $M$ is both left and
right $R$--module with compatible structures.

\begin{defn}\label{d1} \cite[Definition 2.1]{hw}
An $(R, R)$--bimodule $C$ is {\it semidualizing} if it is a finite $R$--module, finite $R^{\mathrm{op}}$--module,
and the following conditions hold.

(1) The homothety map $R\stackrel{^R\gamma}{\longrightarrow}\mathrm{ Hom}_{R^\mathrm{ op}}(C, C)$ is an isomorphism.

(2) The homothety map $R\stackrel{\gamma^R}{\longrightarrow}\mathrm{ Hom}_R(C, C)$ is an isomorphism.

(3) $\mathrm{ Ext}^{\geqslant 1}_R(C, C)=0$.

(4) $\mathrm{ Ext}^{\geqslant 1}_{R^\mathrm{ op}}(C, C)=0$.
\end{defn}
Assume that $R$ is a commutative noetherian ring, then the above definition agrees with the definition of
semidualizing $R$--module (see e.g. \cite[2.1]{hw}).
Also, every finite projective $R$--module of rank 1 is semidualizing (see \cite[Corollary 2.2.5]{s-w}).

\begin{defn} \cite[Definition 3.1]{hw}
A semidualizing $(R, R)$--bimodule $C$ is said to be {\it faithfully semidualizing}
if it satisfies the following conditions

(a) If $\mathrm{Hom}_R(C, M)=0$, then $M=0$ for any $R$--module $M$;

(b) If $\mathrm{Hom}_{R^\mathrm{ op}}(C, N)=0$, then $N=0$ for any $R^\mathrm{ op}$--module $N$.
\end{defn}

Note that over a commutative noetherian ring, all semidualizing modules are faithfully semidualizing,
by \cite[Proposition 3.1]{hw}.

For the remainder of this section $C$ denotes a semidualizing $(R, R)$--bimodule.
The following class of modules is already appeared in, for example, \cite{hj}, \cite{hw}, and \cite{tw}.
\begin{defn}
An $R$--module is called $C$--{\it projective} if it has the form $C\otimes_R P$
for some projective $R$--module $P$. The class of (resp. finite) $C$--projective modules is denoted
by ${\mathcal P}_C$ (resp. ${\mathcal P}^f_C$).
\end{defn}

\begin{f}
A complex $A$ of $R$--modules is called $\mathrm{ Hom}_R(\mathcal{P}_C, -)$--exact
if $\mathrm{ Hom}_R(C\otimes_R P, A)$ is exact for each projective $R$--module $P$.
The term $\mathrm{ Hom}_R(-, \mathcal{P}_C)$--exact is defined dually.
\end{f}

For the notations in the next fact one may see \cite[Definitions 1.4 and 1.5]{wsw1}
\begin{f}\label{d2}
A ${\mathcal P}_C$--{\it resolution} of an $R$--module $M$ is a complex $X$ in ${\mathcal P}_C$
with $X_{-n}=0=\mathrm{H}_n(X)$ for all $n>0$ and $M\cong \mathrm{H}_0(X)$.
The following exact sequence is the {\it augmented} ${\mathcal P}_C$--{\it resolution} of $M$
associated to $X$:
$$X^+=\cdots \stackrel{\partial^X_2}{\longrightarrow} C\otimes_R P_1 \stackrel{\partial^X_1}
{\longrightarrow}C\otimes_R P_0 \longrightarrow M\longrightarrow 0.$$
A ${\mathcal P}_C$--resolution $X$ of $M$ is called $proper$ if in addition $X^+$ is
$\mathrm{Hom}_R({\mathcal P}_C, -)$--exact.

The ${\mathcal P}_C$--{\it projective dimension} of $M$ is the quantity

$\hspace{1.2cm}$ ${\mathcal P}_C$--$\mathrm{pd}(M)=\inf \{\sup \{ n\geqslant0\ |\ X_n\neq0\}\ |\ X$
is an ${\mathcal P}_C$--resolution of $M\}$.

The objects of ${\mathcal P}_C$--projective dimension 0 are exactly $C$--projective $R$--modules.

The notion ({\it proper}) ${\mathcal P}_C$--{\it coresolution} is defined dually.
The {\it augmented} ${\mathcal P}_C$--{\it coresolution} associated to a ${\mathcal P}_C$--coresolution $Y$ is denoted by $^+Y$.
\end{f}

In \cite{tw}, the authors proved the following proposition for a commutative ring $R$. However, by an easy inspection, one can see that it is true even if $R$ is non-commutative.
\begin{prop}\label{p1}
Assume that $C$ is a faithfully semidualizing $(R, R)$--bimodule and that $M$ is an $R$--module.
The following statements hold true.
\begin{itemize}
\item[(a)]\cite[Corollary 2.10(a)]{tw}  The inequality ${\mathcal P}_C$--$\mathrm{pd}(M)\leqslant n$ holds if and only if there is a complex
$$ 0\longrightarrow C\otimes_R P_n\longrightarrow \cdots \longrightarrow C\otimes_R P_0 \longrightarrow M\longrightarrow 0$$
which is $\mathrm{ Hom}_R({\mathcal P}_C,  -)$--exact.
\item[(b)]\cite[Theorem 2.11(a)]{tw}  $\mathrm{ pd}_R(M)=\mathcal{P}_C$--$\mathrm{ pd}_R(C\otimes_R M)$.
\item[(c)] \cite[Theorem 2.11(c)]{tw} $\mathcal{P}_C$--$\mathrm{ pd}_R(M)=\mathrm{ pd}_R(\mathrm{ Hom}_R(C, M))$.
\end{itemize}
\end{prop}

\begin{rem}\label{f1}
By \cite[Proposition 5.3]{hw} the class ${\mathcal P}_C$ is
precovering, that is,  for an $R$--module $M$, there exists a
projective $R$--module $P$ and a homomorphism $\phi:
C\otimes_RP\rightarrow M$ such that, for every projective $Q$, the
induced map
$$\mathrm{ Hom}_R(C\otimes_RQ, C\otimes_RP)\stackrel{\mathrm{ Hom}_R(C\otimes_RQ, \phi)}{-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}\longrightarrow}
\mathrm{ Hom}_R(C\otimes_RQ, M)$$
is surjective. Then one can iteratively take precovers to construct a complex
\begin{equation}\tag{\ref{f1}.1}{W = \cdots\stackrel{{\partial}_2^{X}}{\longrightarrow }
C\otimes_R P_1\stackrel{{\partial}_1^{X}}{\longrightarrow }C\otimes_R P_0\longrightarrow 0}
\end{equation}
such that $W^+$ is $\mathrm{ Hom}_R({\mathcal P}_C,  -)$--exact, where
$$W^+ = \cdots\stackrel{{\partial}_2^{X}}{\longrightarrow }C\otimes_R P_1\stackrel{{\partial}_1^{X}}{\longrightarrow }
C\otimes_R P_0 \stackrel{\phi}{\longrightarrow } M\longrightarrow 0.$$
For the notions precovering, covering, preenveloping and enveloping one can see \cite{ej}.
\end{rem}

Note that if $C$ is faithfully semidualizing $(R, R)$--bimodule and $M$ is an $R$--module,
then, by Proposition \ref{p1}(a), ${\mathcal P}_C$--$\mathrm{pd}(M)$ is equal to the length of
the shortest complex as (\ref{f1}.1).
Thus for any $R$--module $M$, the quantity ${\mathcal P}_C$--projective dimension of $M$,
defined in \cite{hw} and \cite{tw}, is equal to ${\mathcal P}_C$--$\mathrm{pd}(M)$ in \ref{d2}.

\section{Results} \label{sec3}

A ring $R$ is (left) hereditary if every left ideal is projective. The Cartan-Eilenberg theorem \cite[Theorem 4.19]{r}
shows that $R$ is hereditary if and only if every submodule of a projective module is projective.
We show that the quality of being hereditary can be detected by $C$--projective modules, which is interesting on its own.

\begin{prop}\label{H}
Assume that $C$ runs trough the class of faithfully semidualizing $(R, R)$--bimodules.
The following statements are equivalent.
\begin{itemize}
\item[(i)] $R$ is left hereditary.
\item[(ii)] For any $C$, every submodule of a $C$--projective $R$--module is also $C$--projective.
\item[(iii)] There exists a $C$ such that every submodule of a $C$--projective $R$--module is also $C$--projective.
\end{itemize}
\end{prop}

\begin{proof}
(i)$\Rightarrow$(ii). Let $C$ be a faithfully semidualizing bimodule and $N$ a submodule of $C\otimes_R P$, where $P$ is a projective $R$--module.
Then one gets the exact sequence $0 \longrightarrow \mathrm{ Hom}_R(C, N) \longrightarrow P$.
As $R$ is left hereditary, $\mathrm{ Hom}_R(C, N)$ is a projective $R$--module. By Proposition~\ref{p1}(c), $\mathcal{P}_C$--$\mathrm{pd}(N)=\mathrm{pd}(\mathrm{ Hom}_R(C, N))=0$.

(ii)$\Rightarrow$(iii) is immediate.

(iii)$\Rightarrow$(i). As every submodule of a $C$--projective $R$--module is $C$--projective, for any $R$--module $M$
one has $\mathcal{P}_C$--$\mathrm{pd}(M)\leqslant 1$.
Then for any $R$--module $N$ one gets $\mathrm{pd}(N)=\mathcal{P}_C$--$\mathrm{pd}(C\otimes_R N)\leqslant 1$,
by Proposition~\ref{p1}(b). It follows that every submodule of a projective is projective and so,
by \cite[Theorem 4.19]{r}, $R$ is left hereditary.
\end{proof}

\begin{defn}\label{d5}
A complex $X_\bullet$ of $R$--modules is called $C$--{\it perfect} if it is quasiisomorphic to a finite complex
$$T_\bullet= 0 \longrightarrow C\otimes_R P_n\longrightarrow C\otimes_R P_{n-1}\longrightarrow\cdots
\longrightarrow C\otimes_R P_1\longrightarrow C\otimes_R
P_0\longrightarrow 0,$$ where $P_i$ are finite projective
$R$--modules. The {\it width} of such a $C$--perfect complex
$X_\bullet$, denoted by $\mathrm{wd}(X_\bullet)$, is defined to be
the minimal length $n$ of a complex $T_\bullet$ satisfying the above
conditions. A $C$--perfect complex $X_\bullet$ is called {\it indecomposable} if
it is not quasiisomorphic to a direct sum of two non-trivial
$C$--perfect complexes.
\end{defn}

\begin{defn}\label{d6}\cite[Definition 1.1]{bf}
A ring $R$ is called {\it strongly} $r$--{\it regular} if every
perfect complex over $R$ is quasiisomorphic to a direct sum of
perfect complexes of width $\leqslant r$. If $R$ is strongly
$r$--regular for some $r$ then it will be called {\it strongly
regular}.
\end{defn}
\begin{rem}
As Professor Ragnar-Olaf Buchweitz kindly pointed out in his
personal communication with the authors, in \cite{bf} it should be
added the blanket statement that rings are noetherian and modules
are finite. Thus Definition~\ref{d6} agrees with \cite[Definition 1.1]{bf}.
Indeed, over a noetherian ring every perfect complex has bounded and finite homology.
\end{rem}

Note that a hereditary ring $R$ is strongly 1-regular, see \cite[Remark 1.2]{bf}.

In order to bring the results Theorem \ref{J} and Proposition
\ref{p2}, we quote some preliminaries.

\begin{defn}\cite[III.3.2(b)]{gm} and \cite[Definition 2.2.8]{cf}
Let $\alpha: A\rightarrow B$ be a morphism of $R$--complexes. The {\it mapping cone} of $\alpha$,
$\mathrm{Cone}(\alpha)$, is a complex which is given by
$$(\mathrm{Cone}(\alpha))_n=B_n\oplus A_{n-1} \ \ \ \mathrm{and}\ \ \ \ \ \partial_n^{\mathrm{Cone}(\alpha)}=
\left(\begin{array}{cccc}\partial_n^B & \alpha_{n-1}\\ 0 &
-\partial_{n-1}^A\\ \end{array}\right).$$
\end{defn}

It easy to see that the following lemma is also true if $R$ is
non-commutative.
\begin{lem}\label{L10}
Let $\alpha: A\rightarrow B$ be a morphism of $R$--complexes and $M$ be an $R$--module.
The following statements hold true.
\begin{itemize}
\item[(a)] \cite[Lemma 2.2.10]{cf} The morphism $\alpha$ is a quasiisomorphism if and only if $\mathrm{Cone}(\alpha)$ is acyclic.
\item[(b)] \cite[Lemma 2.3.11]{cf} $\mathrm{Cone}(\mathrm{Hom}_R(M, \alpha))\cong \mathrm{Hom}_R(M, \mathrm{Cone}(\alpha))$.
\item[(c)] \cite[Lemma 2.4.11]{cf} $\mathrm{Cone}(M\otimes_R \alpha)\cong M\otimes_R \mathrm{Cone}(\alpha)$.
\end{itemize}
\end{lem}

\begin{rem}\label{R10}
Let $C$ be a semidualizing $(R, R)$--bimodule. Assume that\\
$X=0\rightarrow X_n \rightarrow X_{n-1} \rightarrow \cdots
\rightarrow X_1 \rightarrow X_0 \rightarrow 0$ is an exact complex
of $R$--modules.
\begin{itemize}
\item[(a)] If each $X_i$ is a projective $R$--module, then it is easy to see that the induced complex $C\otimes_R X$ is exact.
\item[(b)] If each $X_i$ is a $C$--projective $R$--module, then the induced complex $\mathrm{Hom}_R(C, X)$ is exact,
since $\mathrm{Ext}_R^{\geqslant 1} (C, X_i)=0$.
\end{itemize}
\end{rem}
\begin{thm}\label{J}
The following statements are equivalent.
\begin{itemize}
\item[(i)] $R$ is strongly $r$--regular.
\item[(ii)] For any faithfully semidualizing bimodule $C$, every $C$--perfect complex is quasiisomorphic
to a direct sum of $C$--perfect complexes of width $\leqslant r$.
\item[(iii)] There exists a faithfully semidualizing bimodule $C$ such that every $C$--perfect complex
is quasiisomorphic to a direct sum of $C$--perfect complexes of width $\leqslant r$.
\end{itemize}
\end{thm}
\begin{proof}
(i)$\Rightarrow$(ii). Let $R$ be strongly $r$--regular, $C$ a
faithfully semidualizing bimodule. Assume that $X_\bullet$ is a
$C$--perfect complex. Then, by Definition~\ref{d5}, there exists a
finite complex $$T_\bullet= 0 \longrightarrow C\otimes_R
P_n\longrightarrow C\otimes_R P_{n-1}\longrightarrow\cdots
\longrightarrow C\otimes_R P_0\longrightarrow 0,$$ such that each
$P_i$ is a finite projective $R$--module and $X_\bullet$ is
quasiisomorphic to $T_\bullet$. Therefore $\mathrm{Hom}_R(C,
T_\bullet)\cong 0 \longrightarrow P_n\longrightarrow
P_{n-1}\longrightarrow\cdots \longrightarrow P_0\longrightarrow 0$
is a perfect complex. By Definition~\ref{d6}, there is a
quasiisomorphism $\alpha:\mathrm{Hom}_R(C,
T_\bullet)\stackrel{\simeq}{\longrightarrow }\bigoplus_{i=1}^s
F_\bullet^{(i)}$, where each $F_\bullet^{(i)}$ is a perfect complex
of width $\leqslant r$. We may assume that each $F_\bullet^{(i)}$ is
a finite complex of finite projective $R$--modules. By
Lemma~\ref{L10}(a), $\mathrm{ Cone}(\alpha)$ is acyclic. As
$\mathrm{Cone}(\alpha)$ is a finite complex of projective
$R$--modules, Remark~\ref{R10} implies that the complex $C\otimes_R
\mathrm{ Cone}(\alpha)$ is acyclic. By Lemma~\ref{L10}, the complex
$\mathrm{ Cone} (C\otimes_R\alpha)$ is acyclic too and so
$C\otimes_R\alpha$ is quasiisomorphism. Therefore $T_\bullet$ is
quasiisomorphic to $\bigoplus_{i=1}^s C\otimes_R F_\bullet^{(i)}$.
Note that each $C\otimes_R F_\bullet^{(i)}$ is a $C$--perfect
complex of width $\leqslant r$.

(ii)$\Rightarrow$(iii) is immediate.

(iii)$\Rightarrow$(i). Let $Y_\bullet$ be a perfect complex. Then,
by Definition~\ref{d5}, there is a finite complex $F_\bullet= 0
\longrightarrow  P_m\longrightarrow  P_{m-1}\longrightarrow\cdots
\longrightarrow P_0\longrightarrow 0$ of finite projective modules
which is quasiisomorphic to $Y_\bullet$. As $C\otimes_R F_\bullet$
is a $C$--perfect complex, our assumption implies that there is a
quasiisomorphism $\beta:C\otimes_R
F_\bullet\stackrel{\simeq}{\longrightarrow }\bigoplus_{i=1}^s
T^{(i)}_\bullet$, where each $T^{(i)}_\bullet$ is a $C$--perfect
complex of width $\leqslant r$. We may assume that, for each $i$,
$$T^{(i)}_\bullet = 0 \longrightarrow C\otimes_R
P_{n_i}^{(i)}\longrightarrow\cdots \longrightarrow C\otimes_R
P_0^{(i)}\longrightarrow 0$$ where each $P_j^{(i)}$ is a finite
projective $R$--module. Similar to the proof of
(i)$\Rightarrow$(ii), one observes that $\mathrm{ Hom}_R(C, \beta)$
is a quasiisomorphism. Therefore $F_\bullet$ is quasiisomorphic to
$\bigoplus_{i=1}^s \mathrm{ Hom}_R(C, T^{(i)}_\bullet)$. Note that
each $\mathrm{ Hom}_R(C, T^{(i)}_\bullet)$ is a perfect complex of
width $\leqslant r$. Thus $R$ is strongly $r$--regular.
\end{proof}

In \cite[Section 1]{am}, Avramov and Martsinkovsky define a general notion of minimality
for complexes: A complex $X$ is {\it minimal} if every homotopy equivalence
$\sigma : X \longrightarrow X$ is an isomorphism.
In \cite[Lemma 4.8]{w}, it is proved that, over a commutative local ring $R$ with maximal ideal $\mathfrak{m}$,
a complex $X$ consisting of modules in $\mathcal P_C^f$ is minimal if and only if $\partial^X(X)\subseteq {\frak m}X$.

In consistent to \cite[Lemma 1.6]{bf} we prove the following
proposition.
\begin{prop}\label{p2}
Let $R$ be a commutative noetherian local ring, $C$ a semidualizing $R$--module. The following
statements hold true.

{\rm (a)} Every $C$--perfect complex $X_\bullet$ is quasiisomorphic to a minimal finite complex
$$T_\bullet= 0 \longrightarrow C\otimes_R F_n\longrightarrow C\otimes_R F_{n-1}\longrightarrow\cdots
\longrightarrow C\otimes_R F_1\longrightarrow C\otimes_R F_0\longrightarrow 0,$$
where each $F_i$ is finite free $R$--module.

{\rm (b)} If two minimal finite complexes of modules of the form $C^m=\oplus^m C$ are quasiisomorphic,
then they are isomorphic.
\end{prop}
\begin{proof}
{\rm (a).} By Definition~\ref{d5}, a $C$--perfect complex $X_\bullet$ is quasiisomorphic to a finite complex
$$T_\bullet= 0 \longrightarrow C\otimes_R P_n\longrightarrow C\otimes_R P_{n-1}\longrightarrow\cdots
\longrightarrow C\otimes_R P_1\longrightarrow C\otimes_R
P_0\longrightarrow 0,$$ where each $P_i$ is a finite free $R$--module.
The complex $\mathrm{Hom}_R(C, T_\bullet)$ is a perfect complex and
so, by \cite[Lemma 1.6(1)]{bf}, there exist a minimal finite complex
$F_\bullet$ of finite free $R$--modules and a quasiisomorphism
$\alpha: \mathrm{Hom}_R(C,
T_\bullet)\stackrel{\simeq}{\longrightarrow } F_\bullet$.
As in the proof of Theorem~\ref{J}, it follows that
$C\otimes_R\alpha: C\otimes_R\mathrm{Hom}_R(C, T_\bullet)\rightarrow C\otimes_R F_\bullet$
is a quasiisomorphism.
As $C\otimes_R F_\bullet$ is a minimal finite complex, we are done.

{\rm (b).} Let $T_\bullet$ and $L_\bullet$ be two minimal finite
complexes of modules of the form $C^m$. Assume that
$\alpha:T_\bullet\rightarrow L_\bullet$ is a quasiisomorphism. Then,
by Remark~\ref{R10} and Lemma~\ref{L10}, $\mathrm{ Hom}_R(C,
\alpha): \mathrm{ Hom}_R(C, T_\bullet)\rightarrow \mathrm{ Hom}_R(C,
L_\bullet)$ is a quasiisomorphism of minimal finite complexes of
finite free $R$--modules. Thus, by the proof of \cite[Lemma
1.6(2)]{bf}, $\mathrm{ Hom}_R(C, \alpha)$ is an isomorphism. Now,
there is a commutative diagram of complexes and morphisms
$$\begin{array}{ccc}
 T_\bullet & \stackrel{\simeq}{\underset{\alpha}{-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}
 -\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}
 \longrightarrow}} & L_\bullet  \\
  \begin{array}{ll} \ \ {\Big\uparrow} \cong \end{array} &     &\begin{array}{ll} \ \ {\Big\uparrow} \cong \end{array}  \\
 C\otimes_R \mathrm{ Hom}_R(C, T_\bullet)
 & \stackrel{\cong}{\underset{C\otimes_R \mathrm{ Hom}_R(C, \alpha)}{-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}
 -\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}\longrightarrow}} &
  C\otimes_R \mathrm{ Hom}_R(C, L_\bullet), \\
\end{array}$$
where the vertical morphisms are natural isomorphisms. This implies
that $\alpha$ itself must be an isomorphism.
\end{proof}

It is proved in \cite[Lemma 4.9]{w} that every finite module $M$ over a commutative noetherian local ring $R$ with
$\mathcal P^f_C$--$\mathrm{ pd}(M)<\infty$ admits a minimal $\mathcal P_C^f$--resolution.
Now we show that every finite $R$--module which has a proper $\mathcal P_C$--resolution, admits a minimal proper one.
Note that if $\mathcal P^f_C$--$\mathrm{ pd}(M)<\infty$ then $M$ admits a proper $\mathcal P_C$--resolution
(see proof of \cite[Corollary 2.10]{tw}).

\begin{thm}\label{E}
Assume that $R$ is a commutative noetherian local ring and that $C$ is a semidualizing $R$--module. Then $\mathcal P_C^f$ is covering in the category of finite $R$--modules. For any finite $R$--module $M$, there is a complex
$X=\cdots \longrightarrow C^{n_1}\longrightarrow C^{n_0}\longrightarrow 0$ with the following properties.

$\mathrm{ (1)}$ $X^+=\cdots \longrightarrow C^{n_1}\longrightarrow C^{n_0}\longrightarrow M \longrightarrow 0$ is
$\mathrm{ Hom}_R(\mathcal P_C, -)$--exact.

$\mathrm{ (2)}$ $X$ is a minimal complex.

If $M$ admits a proper $\mathcal P_C$--resolution, then $X^+$ is exact and so $X$ is a minimal proper $\mathcal P_C$--resolution of $M$.
\end{thm}
\begin{proof}
Let $M$ be a finite $R$--module. Assume that $n_0=\nu (\mathrm{ Hom}_R(C, M))$ denotes
the number of a minimal set of generators of $\mathrm{ Hom}_R(C, M)$
and that $\alpha:R^{n_0}\longrightarrow  \mathrm{ Hom}_R(C, M)$ is the natural epimorphism.
As $\alpha$ is a $\mathcal P^f$--cover of $\mathrm{ Hom}_R(C, M)$, the natural map
$\beta=C\otimes_RR^{n_0}\stackrel{C\otimes_R\alpha}{-\hspace{-0.07cm}-\hspace{-0.17cm}\longrightarrow}
C\otimes_R\mathrm{ Hom}_R(C, M) \stackrel{\nu_M}{\longrightarrow}M$
is a $\mathcal P_C^f$--cover of $M$.
Set $M_1=\mathrm{ Ker} \beta$ and $n_1=\nu(\mathrm{ Hom}_R(C, M_1))$. Thus there is a $\mathcal P_C^f$--cover
$\beta_1: C\otimes_RR^{n_1}\longrightarrow M_1$.
Proceeding in this way one obtains a complex
$$X=\cdots \stackrel{\partial_2=\epsilon_2 \beta_2}{-\hspace{-0.07cm}-\hspace{-0.17cm}\longrightarrow} C\otimes_RR^{n_1}\stackrel{\partial_1=\epsilon_1 \beta_1}{-\hspace{-0.07cm}-\hspace{-0.17cm}\longrightarrow} C\otimes_RR^{n_0}\longrightarrow 0,$$
where $\epsilon_i: M_i \rightarrow C\otimes_RR^{n_{i-1}}$ is the inclusion map for all $i\geqslant 1$.
As the maps in $X$ are obtained by $\mathcal P_C^f$--covers, the complex $X^+$ is $\mathrm{ Hom}_R(\mathcal P_C, -)$--exact.
It is easy to see that $\mathrm{ Hom}_R(C, X)$ is minimal free resolution of $\mathrm{ Hom}_R(C, M)$.
Now we show that $X$ is a minimal complex. Let $f: X\rightarrow X$ be a morphism which is homotopic to $\mathrm{id}_X$.
It is easy to see that the morphism $\mathrm{Hom}_R(C, f)$ is homotopic to $\mathrm{id}_{\mathrm{Hom}_R(C, X)}$.
As the complex $\mathrm{ Hom}_R(C, X)$ is minimal, by \cite[Proposition 1.7]{am}, the morphism $\mathrm{Hom}_R(C, f)$ is an isomorphism.
The commutative diagram
$$\begin{array}{ccc}
 X & \stackrel{f}{{-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}
 -\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}
 \longrightarrow}} & X  \\
  \begin{array}{ll} \ \ {\Big\downarrow} \cong \end{array} &     &\begin{array}{ll} \ \ {\Big\downarrow} \cong \end{array}  \\
 C\otimes_R \mathrm{ Hom}_R(C, X)
 & \stackrel{\cong}{\underset{C\otimes_R \mathrm{ Hom}_R(C, f)}{-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}
 -\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}-\hspace{-0.2cm}\longrightarrow}} &
  C\otimes_R \mathrm{ Hom}_R(C, X), \\
\end{array}$$
with vertical natural isomorphisms, implies that $f$ is an
isomorphism. Therefore, by \cite[Proposition 1.7]{am}, $X$ is
minimal. If $M$ admits a proper $\mathcal P_C$--resolution, then by
\cite[Corollary 2.3]{tw}, $X^+$ is exact.
\end{proof}
The proof of the next lemma is similar to \cite[Corollary 2.3]{tw}.

\begin{lem}\label{g}
Let $R$ be a commutative noetherian ring and let $M$ be a finite $R$--module. Assume that $C$ is a semidualizing $R$--module.
The following are equivalent.
\begin{itemize}
\item[(i)] $M$ admits a proper $\mathcal P^f_C$--coresolution.
\item[(ii)]  Every $\mathrm{ Hom}_R(-, \mathcal P^f_C)$--exact complex of the form
$$0\longrightarrow M \longrightarrow C\otimes_RQ_0 \longrightarrow C\otimes_RQ_{-1}\longrightarrow \cdots$$
is exact, where $Q_i$ is an object of $\mathcal P^f$ for all $i\leqslant0$.
\item[(iii)] The natural homomorphism $M \longrightarrow \mathrm{ Hom}_R(\mathrm{ Hom}_R(M, C), C)$ is an isomorphism and $\mathrm{ Ext}^{\geqslant1}_R(\mathrm{ Hom}_R(M, C), C)=0$.
\end{itemize}
\end{lem}

\begin{prop}\label{F}
Assume that $R$ is a commutative noetherian local ring and that $C$ is a semidualizing $R$--module. Then $\mathcal P_C^f$ is enveloping in the category of finite $R$--modules. For any finite $R$--module $M$, there is a complex
$Y=0 \longrightarrow C^{m_0}\longrightarrow C^{m_1}\longrightarrow \cdots$ with the following properties.

$\mathrm{ (1)}$ $ ^+Y=0\longrightarrow M\longrightarrow  C^{m_0}\longrightarrow  C^{m_1} \longrightarrow \cdots $ is $\mathrm{ Hom}_R(-, \mathcal P_C)$--exact.

$\mathrm{ (2)}$ $Y$ is a minimal complex.

If $M$ admits a proper $\mathcal P^f_C$--coresolution, then $^+Y$ is exact and so $Y$ is a minimal proper $\mathcal P_C$--coresolution of $M$.
\end{prop}
\begin{proof}
Let $M$ be a finite $R$--module. Assume that $m_0=\nu (\mathrm{
Hom}_R(M, C))$ denotes the number of a minimal set of generators of
$\mathrm{ Hom}_R(M, C)$ and that $\alpha:R^{m_0}\longrightarrow
\mathrm{ Hom}_R(M, C)$ is the natural $\mathcal P^f$--cover of
$\mathrm{ Hom}_R(M, C)$. It follows that
$\gamma=M\stackrel{\delta_M}{\longrightarrow}\mathrm{
Hom}_R(\mathrm{ Hom}_R(M, C), C)\stackrel{\mathrm{ Hom}_R(\alpha,
C)}{-\hspace{-0.07cm}-\hspace{-0.17cm}\longrightarrow}\mathrm{
Hom}_R(R^{m_0}, C)$ is a $\mathcal P_C^f$--envelope of $M$. Set
$M_{-1}=\mathrm{ Coker} \gamma$ and $m_1=\nu (\mathrm{
Hom}_R(M_{-1}, C))$. As mentioned, there is a $\mathcal
P_C^f$--envelope $\gamma_1:M_{-1}\longrightarrow\mathrm{
Hom}_R(R^{m_1}, C)$. Proceeding in this way one obtains a complex
$Y=0 \longrightarrow \mathrm{ Hom}_R(R^{m_0},
C)\stackrel{\partial_0=\gamma_1\pi_1}{-\hspace{-0.07cm}-\hspace{-0.17cm}\longrightarrow}
\mathrm{ Hom}_R(R^{m_1},
C)\stackrel{\partial_{-1}=\gamma_2\pi_2}{-\hspace{-0.07cm}-\hspace{-0.17cm}\longrightarrow}
\cdots$, where $\pi_i$ is the natural epimorphism for all
$i\geqslant 1$. Since the maps in $Y$ are obtained by $\mathcal
P_C^f$--envelopes, the complex $ ^+Y$ is  $\mathrm{ Hom}_R(-,
\mathcal P_C)$--exact. It is easy to see that $\mathrm{ Hom}_R(Y, C)$
is minimal free resolution of $\mathrm{ Hom}_R(M, C)$.
Similar to the proof of Theorem~\ref{E}, we find that $Y$ is a minimal complex.
If $M$ admits a proper $\mathcal P^f_C$--coresolution,
then, by Lemma \ref{g}, $^+Y$ is exact.
\end{proof}

In the following example we find an $R$--module $M$ with $\mathcal
P_C$--$\mathrm{ pd}(M)=\infty$ which admits a minimal proper
$\mathcal P_C$--resolution. This example shows that a commutative
noetherian local ring which admits an exact zero-divisor is not a
strongly regular ring.

\begin{exmp}\label{ex3}
Let $R$ be a commutative noetherian local ring, $C$ a semidualizing $R$--module.
Assume that $x,  y$ form a pair of exact zero-divisors on both $R$ and $C$ (e.g. see \cite[Example 3.2]{ad}).
Then $\mathcal P_C$--$\mathrm{ pd}(C/xC)=\mathrm{ pd}(R/xR)=\infty$.  The complex
$$T_\bullet=\cdots \stackrel{x}{\longrightarrow}C \stackrel{y}{\longrightarrow}C\stackrel{x}{\longrightarrow}C\longrightarrow 0\ (\ \mathrm{ resp.}\ L_\bullet= 0\longrightarrow C \stackrel{x}{\longrightarrow}C \stackrel{y}{\longrightarrow}C\stackrel{x}{\longrightarrow}\cdots )$$
is a minimal $\mathcal P_C$--resolution (resp. $\mathcal P_C$--coresolution) of $C/{xC}$.
By \cite[Proposition 3.4]{ad}, $C/xC$ is a semidualizing $R/xR$--module. By \cite[Proposition 2.13]{dg}, there are isomorphisms
$$\mathrm{ Hom}_R(C, C/{xC})\cong \mathrm{ Hom}_{R/xR}(C/{xC}, C/{xC})\cong R/xR,$$
$$\mathrm{ Hom}_R(C/{xC}, C)\cong \mathrm{ Hom}_{R/xR}(C/{xC}, C/{xC})\cong R/xR.$$
Applying $\mathrm{ Hom}_R(C, -)$ and $\mathrm{ Hom}_R(-, C)$ on the above complexes, respectively, would result the isomorphisms $\mathrm{ Hom}_R(C, T_\bullet^+)\cong F_\bullet^+$ and $\mathrm{ Hom}_R( ^+L_\bullet, C)\cong F_\bullet^+$, where
$F_\bullet^+$ is the exact complex
$\cdots \stackrel{y}{\longrightarrow}R\stackrel{x}{\longrightarrow}R \stackrel{y}{\longrightarrow}R\stackrel{x}{\longrightarrow}R\longrightarrow R/xR\longrightarrow 0.$
Therefore $T_\bullet$ (resp. $L_\bullet$) is a minimal proper $\mathcal P_C$--resolution (resp. $\mathcal P_C$--coresolution) of $C/{xC}$.

For each $n$, one obtains a $C$--perfect complex of length $n$ as
$$T_\bullet^{(n)}= 0 \longrightarrow C {\longrightarrow}C {\longrightarrow}\cdots\stackrel{x}{\longrightarrow}C\stackrel{y}{\longrightarrow}C\stackrel{x}{\longrightarrow}C \longrightarrow 0,$$
where $T_i^{(n)}= T_i$ for all $0\leq i \leq n$ and $T_i^{(n)}=0$ otherwise.
Note that the induced map $\bar{d}_i:T_i^{(n)}/{\mathrm{ Ker}\, d_i}\rightarrow T_{i-1}^{(n)}$ is injective,
where $\mathrm{ Ker}\, d_i$ is equal to $yC$ or $xC$. As $C$ is indecomposable $R$--module, $T_\bullet^{(n)}$ is indecomposable which has a similar proof to \cite[Proposition 1.5]{bf}.
\end{exmp}

\section*{\bf Acknowledgment}
The authors are grateful to the referee for his/her careful reading
of the paper and valuable comments. The second author was supported
in part by a grant from IPM (No.93130110).

redpajama_set_name: RedPajamaArXiv


GARCIA-FERNANDEZ, M.; LIMA, Marcos Vinicius Borges Teixeira. Weak lensing magnification in the dark energy survey science verification data. Monthly Notices of the Royal Astronomical Society, Oxford, v. 476, n. 1, p. 1071-1085, 2018. DOI: 10.1093/mnras/sty282.
Garcia-Fernandez M, Lima MVBT. Weak lensing magnification in the dark energy survey science verification data. Monthly Notices of the Royal Astronomical Society. 2018 ; 476( 1): 1071-1085.
Título do periódico: Annu. Rev. Nucl. Part. Sci.
Título do periódico: J. Am. Stat. Assoc.
Título do periódico: J. Mach. Learn. Res.


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On December 20, 2019, the president signed into law the National Defense Authorization Act for Fiscal Year 2020, which established the Liberian Refugee Immigration Fairness ("LRIF") program. LRIF is the first US legalization program – creating a path to lawful permanent resident (LPR) status – in many years. It applies to select Liberian nationals – most of them long-term US residents who qualified for Temporary Protected Status (TPS) and Deferred Enforced Departure (DED) based on the insecure conditions in Liberia – and their close family members.
To qualify, Liberian nationals must:
Apply by December 20, 2020;
Have been continuously present in the United States since November 20, 2014;
Be admissible or eligible for a waiver of inadmissibility or another form of relief;
Have not been convicted of an aggravated felony or two or more crimes of moral turpitude; and
Have not persecuted others on account of race, religion, nationality, membership in a particular social group, or political opinion.
Select family members are eligible if they:
Are the spouse, an unmarried child under age 21, or an unmarried child 21 years or older of a qualifying Liberian national;
Are admissible, or eligible for a waiver of inadmissibility or another form of relief;
There is a significant risk that many eligible Liberians and their family members may not meet the application deadline due to the effects of the COVID-19 pandemic and poor roll-out of the program. To highlight this concern, the Center for Migration Studies of New York (CMS) has produced estimates – rounded to the nearest hundred – of the Liberian nationals who arrived in 2014 or earlier, and who are not naturalized US citizens or LPRs, and of their non-US citizen, non-LPR spouses and unmarried children who are also potentially eligible to adjust under LRIF. CMS has also estimated the number of essential workers among LRIF-eligible Liberian nationals, as well as their median personal and household incomes. In particular, it estimates that:
10,000 Liberian nationals are potentially eligible for LRIF, including 7,900 US critical essential infrastructure workers;
The median personal income of potential LRIF beneficiaries is $23,000 and the median household income is $46,000; and
An additional 100 spouses and 200 children of potential LRIF beneficiaries are potentially eligible as the family members of LRIF beneficiaries (Appendix A).
These estimates, of course, include persons that may be inadmissible or who are otherwise barred from the program.
To arrive at these estimates, CMS used 2018 1-Year American Community Survey (ACS) data to identify Liberians without US citizenship or LPR status that arrived in the United States in 2014 or earlier. It then identified their spouses and unmarried children without citizenship or LPR status. These estimates only include spouses and children that lived with Liberians in 2018. All figures are rounded to the nearest 100.
CMS imputed LPR and citizenship status using 2018 1-Year ACS data. A description of CMS estimation procedures, as well as a discussion of the plausibility of the estimates, is provided in Warren (2020).
CMS's estimates of essential workers encompass all persons in the labor force aged 16 and over who were employed in "essential critical infrastructure" categories, as defined by the US Department of Homeland Security, in 2018. A detailed description of CMS's methodology for estimating essential workers is provided in Kerwin et al. 2020.
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Schumann described this opus as a sort of adventure story in instalments; and in a sense it does use serial techniques. As in most of his large-scale works, there is a motto-theme, of which the abstract notes (like purloined letters) can be hidden in a way. too obvious to he noticeable.
I find Dino Ciani himself a shade too reticent.
Florestan, it will be remembered, leaped on the piano to make a speech: that is what the music does more often than the pianist in this performance. No doubt it is difficult to believe a composer who says he is telling stories. But op 21 is declaredly programmatic, with overtones of knight-and-maiden chivalry and other homage to Clara added to overt literary associations with Goethe and Shakespeare. However, in a young pianist making his record debut the absence of grandiloquence is a virtue. The music is played with true dedication, and one of its main aspects is captured fìnely. At 12s 6d the disc would be a real bargain even if it had a competitor; as the only complete set of Novelletten in the catalogue. It is bound to appear on the shelves of every Schumann-lover.


text:
Q: String.Format (format,date) is ignoring format Not sure what's going on here.
I have a DateTime object, and when I try:
String.Format( "{0:dd/MM/yyyy}", _date)

the value returned is:
"24-05-1967"

What I want is
"24/05/1967"

Can anyone explain why my format string is being ignored?
A bit more background:
This is a web app which started out life as .net 1.1, and I'm in the process of moving it up to 2.0/3.5.
Update:
If I change the format to {0:dd:MM:yyyy}, it returns 24:05:1967 - it's only the / in the format string that gets changed to the - char.

Resolution:
When updating the app to run under 2.0, the asp.net globalization settings were messed up.
From the web site properties, ASP.NET tab, Edit Configuration, Application Tab - the culture and UI Culture were both set to the first item in the list (af-ZA) for some bizarre reason.

A: The / is actually the date separator for your specific culture which could be -, in other words, the format string is not ignored but actually used correctly. Look at what CultureInfo is associated with the running thread:
System.Threading.Thread.CurrentThread.CurrentCulture

If you try this:
String.Format(new CultureInfo("en-US"), "{0:dd/MM/yyyy}", DateTime.Now);

You will get the date formatted with / since that is the correct separator for en-US. What you probably should do is use the short date format string and make sure that the thread has the appropriate culture associated, then this would get you what you want and it will work in a globalized application as well:
DateTime.Now.ToString("d", Thread.CurrentThread.CurrentCulture);

Hope this helps!

A: You likely want to use the ToString() method of your DateTime object to get the string representation you are after. A format string of "dd'/'MM'/'yyyy" will give you the format you described. Note that you need to escape literal characters in the format string using single quotes.
Example: DateTime.Now.ToString("dd'/'MM'/'yyyy");

A: Unless you specify the IFormatProvider (usually a CultureInfo object), it will by default use Thread.CurrentThread.CurrentCulture's date/time formatting, which will give you the changing results.
And yes, it will change / but not :.
The solution in this case is to specify the format provider, like this:
String.Format(CultureInfo.InvariantCulture, "{0:dd/MM/yyyy}", _date)

A: The "/" character is the culture neutral placeholder for the date time separator character which is defined in Regional Settings. So is the ":" character for time.
To embed the "/" literally and not as a placeholder you need to enclose it in single quotation marks '/' in the format string.

A: That's odd.  That formatting works correctly for me.  You might want to try _date.ToString("dd/MM/yyyy") instead (though that's just a shot in the dark).

A: Regional settings (on web server?) have date delimiters set to '-'. In my locale (EE) output would be "24.05.1967" :)

A: This will give you your required output
String.Format( "{0:dd/MM/yyyy}", Convert.ToDateTime(_date));

A: Interesting thread !
I have an application which reads in an Excel file, and lets the user export it's data to a .csv file in a user-defined format.
I specifically wanted to give the users the ability to export dates to (exactly) a format like yyyy/mm/dd, regardless of their laptop's culture info.
I tried a few of these suggestions, but the only one which did work was Pop Catalin's suggestion, to wrap any / characters with apostrophes:
outputFormat = outputFormat.Replace("/", "'/'");
valueToExport = ImportedDate.ToString(outputFormat);

This seems to be the only way to force ToString() to use the exact format string, without trying to do anything Culture specific.
Update
I always find it amusing when I answer a StackOverflow question, then hit the same bug a few years later, and stumble across my own answer !
As mentioned in lots of these answers, the solution is to wrap forward-slash characters in apostrophes, to force ToString() to ignore the culture settings.
So, if your Culture settings have a full-stop character as the date separator character (eg "20.07.2015"), then here's what you'd see, when attempting to nicely format the date of Christmas Day, and how you can easily force it to always use forward-slashes:
DateTime dtChristmas = new DateTime(2015, 12, 31);

//  This might return "31/12/2015", "31.12.2015", etc, depending on Culture settings
string str1 = dtChristmas.ToString("dd/MM/yyyy");

//  ...but these two statements will both *always* return "31/12/2015"
string str2 = dtChristmas.ToString("dd'/'MM'/'yyyy");
string str3 = dtChristmas.ToString("dd/MM/yyyy", System.Globalization.CultureInfo.InvariantCulture);

redpajama_set_name: RedPajamaStackExchange


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text:
L'acronyme FNAM peut faire référence à :
 Fédération nationale André Maginot, association française d'utilité publique d'aide aux anciens combattants ;
Fédération nationale de l'aviation et de ses métiers, principale organisation professionnelle dans le transport aérien français.

Articles connexes
 Fédération nationale des agriculteurs multiplicateurs de semences (Fnams)
Fédération nationale des associations d'accueil de marins (FNAAM), fédération française membre de l'

redpajama_set_name: RedPajamaWikipedia


As a private practice clinic owner and as a member of our community, you're might be wondering if a strategy session with the Clinic Mastery team could help you move forward with greater clarity and momentum.
Being in business is tough, the daily headaches of operations can often make you feel like you're always putting out spot fires, taking your attention away from the important business growth activities.
If you feel like you're lacking control of what's going on week to week, then you'd probably benefit from getting some clarity and defining some structure with a strategy session moving forward.
Perhaps you're even wondering if the Business Academy is appropriate for you.
If you are serious about working together and receiving the support you need to go to the next level, then the first and most appropriate step would be for us to meet one another as part of a strategy session.
After working with 100s of private practice owners and as part of our Business Academy, we know that having clarity about your situation, structure moving forward and accountability to implement the changes necessary often requires another perspective. That's what we will give you at a strategy session.
Why Does Clinic Mastery Offer Strategy Sessions?
It's getting quite the reputation for being the most practical, no fluff, actionable business training for clinic owners!
It really speaks to our 'style' of, let's pull up our sleeves and focus on getting results.
The thing is, we've got a lot to offer, but at the end of the day, it's you, the clinic owner who has to implement everything on the front line of your business each day.
What that means is that we need to find out who's cut out for the intensity of the academy, who's got the hunger and who's passionate about transforming the way they lead their clinic.
The best way for us to determine who could join us in one of the 4 enrolments we have each year is to meet you before you apply.
Now, you're time is valuable so we thought that if you are serious about growing your clinic, we'd be willing to give you a heap of value upfront to show you what you could get as a member.
If we're not a good fit after the strategy session, then we can go our separate ways and you'll have at least moved on with some clear action points to implement.
If we are a good fit, then we'll definitely follow up with a second strategy session to review and progress to the application stage.
Who is it For, Is it Right for You?
The strategy session is ideal for someone who looking to join our Business Academy and really take their clinic to the next level.
The Clinic Mastery Business Academy is an Exclusive Business Accelerator that is by Application Only for clinic owners who are dissatisfied with the rate they are growing and the lack of support and direction available for them.
You have a grand vision for transforming your client's experience, you're not bound by tradition.
You're driven by practicality and implementation, so you enjoy taking on a new challenge.
If you believe that you have these key attributes, and you're considering the Business Academy, then let's start with a Strategy session.
Have clarity on underlying cause of your businesses current reality, position and situation.
Have structure in the form of 1-3 actionable steps to implement in order to move the business forward with purpose and precision.
Have an opportunity to learn more and apply for the Business Academy if suitable.
How is The Strategy Session Structured?
Prior to your strategy session, you will be asked to complete a short business audit which is a series of questions to give us insight into your clinic.
It's important that you answer the questionnaire thoroughly because if you don't our analysis of your business will be off, which means we won't know how to help you navigate the growth ahead.
The session will ideally take place on an online video call, either Skype or Zoom, but we'll have a phone as back up. That's so that we can meet face to face and make a personal connection.
Well explore where you clinic is at right now by reviewing your key statistics and also getting further context on your challenges.
We'll identify the highest, most important and valuable aspect of your business that you need to transform and provide 1-3 key actionable action you must take.
At that point if you see the value in working together, and we believe you'd be a good fit, great, we'll organise a second strategy session.
At the second session we can review your ability to implement, refine your plan and if appropriate, you'll be able to Apply for the next Business Academy enrolment.
If you don't want to move forward, that's fine too. Worst case you would have received some free advice from the Clinic Mastery team which will help you generate growth into the future.
Your initial strategy session will be up to 45 minutes.
This is where we will figure out exactly what you need, and how to make it happen.
All you need to do is invest your time, 45 minutes.
You're time is valuable so we thought that if you are serious about growing your clinic, we'd be willing to give you a heap of value upfront to show you what you could get as a member.
If you're going to make and investment of time and money, you want to be confident that it is going to be worthwhile.
So, what we can do is provide you with a minimum of 1 Strategy Session to help you find comfort in your decision and see value in your investment.
What are the Obligations & Expectations of You?
Throughout the process, we want to be judged with whether we live by what we say, because, any time we don't, we don't believe we deserve to retain you as a member of our community.
You should expect us to have your best interests far and way ahead of ours. You should never worry or concern that we are doing anything just for the economic purpose above your interests.
We'll help you take control of your situation with least amount of stress and the most amount of clarity.
If we notice prior to the strategy session that you're not suitable or a conflict exists within our community, we'll tell you beforehand and direct you elsewhere.
We'll focus internally on clinic inefficiencies so you minimise expenditure and have more control over the results.
We'll deliver our recommendation within the context of Leading Great Teams to Deliver Amazing Client Experiences and put People before Profits.
If we believe you are an exceptional representation of what it means to be a Clinic Mastery Member, we will invite you a second Strategy session.
What Happens After the First Strategy Session?
If at the end of the first strategy session we believe your clinic is not suitable for us to continue working with, then we will advise you.
You will have taken some good strategies and insights how you can grow your clinic.
If at the end of the first strategy session you believe we have given you value and we believe that there's a chance we could work together, then we'll organise a second strategy session.
At the second strategy session we'll review you ability to implement change in your clinic by what you've done since the first session and determine if the business academy is right for you.
If you are interested and once you are approved by a Clinic Mastery Team Member, you will be invited to apply to enrol in the next (only 4 p.a.) Business Academy cohort.
If you're serious about growing your clinic, let's talk!
There is only upside in having a practical strategic session.
This is not pitch fest or sales call.
Even if you are going to join the Academy, we want the strategy session to be massively practical because you're time is so valuable.
The reality is that we only have a handful of clinic's join us at each one of the 4 enrolments per year and we're only accepting committed clinic owners.
It's likely you have worked hard to get to this point, but you'd like to experience more fulfilment and reap greater rewards for being in business.
If you think we could help you experience that, then please organise a time today!


text:
Q: Mopub Banner ads is not appearing I am kind of newbie in Android Programming, So I Just want to Display Mopub banner ad in my app. I created the Sample app, also created Ad unit id in Mopub.
I successfully integrated SDK and following their https://developers.mopub.com/docs/android/banner/ instructions on this page.
but I don't know why my app which is Live on G-play Only sending the request but no impression. I also checked on my Phone & its true. No ads are displaying. Here is My Code
Can anyone Help me Please?
mainactivity.java:-
import com.mopub.mobileads.MoPubErrorCode;
import com.mopub.mobileads.MoPubView;
import static com.mopub.mobileads.MoPubView.BannerAdListener;

public class MainActivity extends AppCompatActivity implements BannerAdListener {
private MoPubView mBanner;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    mBanner = (MoPubView) findViewById(R.id.bannerview);
    mBanner.setAdUnitId("XXXXXXXXXXXXXXXXXXXXXXXX"); // Enter your Ad Unit ID from www.mopub.com
    mBanner.setBannerAdListener(this);
    mBanner.loadAd();

}

@Override
protected void onDestroy() {
    mBanner.destroy();
    super.onDestroy();
}

@Override
public void onBannerLoaded(MoPubView banner) {
     Log.d("MoPub Demo", "Banner loaded callback.");
}

@Override
public void onBannerFailed(MoPubView banner, MoPubErrorCode errorCode) {
    Log.d("MoPub Demo", "Banner failed callback with: " + errorCode.toString());
}

@Override
public void onBannerClicked(MoPubView banner) {
    Log.d("MoPub Demo", "Banner clicked callback.");
}

@Override
public void onBannerExpanded(MoPubView banner) {
    Log.d("MoPub Demo", "Banner expanded callback.");
}

@Override
public void onBannerCollapsed(MoPubView banner) {
    Log.d("MoPub Demo", "Banner collapsed callback.");
}
}

Activity_main.xml:-
<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout
xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:app="http://schemas.android.com/apk/res-auto"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
tools:context=".MainActivity">

<TextView
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:text="Hello World!"
    app:layout_constraintBottom_toBottomOf="parent"
    app:layout_constraintLeft_toLeftOf="parent"
    app:layout_constraintRight_toRightOf="parent"
    app:layout_constraintTop_toTopOf="parent" />

<com.mopub.mobileads.MoPubView
    android:id="@+id/bannerview"
    android:layout_width="fill_parent"
    android:layout_height="50dp"
    android:layout_alignParentBottom="true"
    />

</RelativeLayout>

Build.gradle(module.app):-
repositories {
      // ... other project repositories
         jcenter() // includes the MoPub SDK and AVID library
            maven { url "https://s3.amazonaws.com/moat-sdk-builds" }
         }
apply plugin: 'com.android.application'

android {
      compileSdkVersion 27
       defaultConfig {
    applicationId "com.your.prject.appmopubtest"
    minSdkVersion 16
    targetSdkVersion 27
    versionCode 1
    versionName "1.0"
    testInstrumentationRunner "android.support.test.runner.AndroidJUnitRunner"
}
buildTypes {
    release {
        minifyEnabled false
        proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
    }
}
}

dependencies {
implementation fileTree(dir: 'libs', include: ['*.jar'])
implementation 'com.android.support:appcompat-v7:27.1.1'
implementation 'com.android.support.constraint:constraint-layout:1.0.2'
implementation 'com.google.android.gms:play-services-ads:12.0.1'
implementation 'com.google.android.gms:play-services-auth:12.0.1'
testImplementation 'junit:junit:4.12'
androidTestImplementation 'com.android.support.test:runner:1.0.1'
androidTestImplementation 'com.android.support.test.espresso:espresso-core:3.0.1'
implementation('com.mopub:mopub-sdk:4.20.0@aar') {
    transitive = true
}
}

Build.gradle(project.appmopubtest):-
// Top-level build file where you can add configuration options common to all sub-projects/modules.

buildscript {

repositories {
    google()
    jcenter()
    maven { url "https://s3.amazonaws.com/moat-sdk-builds" }

}
dependencies {
    classpath 'com.android.tools.build:gradle:3.1.0'

    // NOTE: Do not place your application dependencies here; they belong
    // in the individual module build.gradle files
}
}

allprojects {
repositories {
    google()
    jcenter()
}
}

task clean(type: Delete) {
delete rootProject.buildDir
}

androidmanifest.xml:-
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.your.prject.appmopubtest">
<!-- Required permissions -->
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

<!-- Optional permissions. Will pass Lat/Lon values when available. Choose either Coarse or Fine -->
<uses-permission android:name="android.permission.ACCESS_COARSE_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION"/>
<application
    android:allowBackup="true"
    android:icon="@mipmap/ic_launcher"
    android:label="@string/app_name"
    android:roundIcon="@mipmap/ic_launcher_round"
    android:supportsRtl="true"
    android:theme="@style/AppTheme">
    <activity android:name=".MainActivity">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>

    <!-- All ad formats -->
        <activity android:name="com.mopub.common.MoPubBrowser" android:configChanges="keyboardHidden|orientation|screenSize"/>

    <!-- Interstitials -->
    <activity android:name="com.mopub.mobileads.MoPubActivity" android:configChanges="keyboardHidden|orientation|screenSize"/>
    <activity android:name="com.mopub.mobileads.MraidActivity" android:configChanges="keyboardHidden|orientation|screenSize"/>

    <!-- Rewarded Video and Rewarded Playables -->
    <activity android:name="com.mopub.mobileads.RewardedMraidActivity" android:configChanges="keyboardHidden|orientation|screenSize"/>

    <activity android:name="com.mopub.mobileads.MraidVideoPlayerActivity" android:configChanges="keyboardHidden|orientation|screenSize"/>

</application>

</manifest>


Previous message (by thread): [Snort-users] gigabit nic?
Next message (by thread): [Snort-users] gigabit nic?
(1000BT/copper) I've ended up using Intel Pro/1000 MF NICs (1000B-SX/fibre).
support a 33MHz PCI bus.
Adaptec.  File systems is using LVM.
> 250Mbits, and Snort doesn't barf and I don't show dropped packets.
Running one snort process and one barnyard process (alert).


Looking for a new home in Gold Run? It is a great time to buy real estate in Gold Run and our site is a great place to begin your search. You can check out Gold Run schools, market statistics, the latest listings and more. Whether you are a first-time home buyer or you are already familiar with the home buying process, you can be assured that you have the best tools and the perfect advisor available to help with your Gold Run home search.
There are currently 5 lots/acres/farmss for sale in Gold Run. The average price for Gold Run properties is $32,860which is 84% lower than all lots/acres/farmss within a 20 mile radius. The highest priced property in Gold Run is $36,500 while the lowest priced lots/acres/farms is listed at $29,900. Most of the properties available in Gold Run are lots/acres/farmss . Compared to last month, the average price of real estate in the area has gone up 100%.


Great!!! With the flexible butterfly hair clip you can easily and quickly make a new hairstyle in 30 seconds and give you a new experience and a special feeling.
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We mainly focus on discovering the most innovative products, gadgets, and tools bringing conveniences for our lives, making lives easier and more comfortable!
If you have any questions about our products, please contact us online or mail to: homieshack@dtcreative.zendesk.com!


text:
Home>News Center>China
HK Chief Executive Tung offers to resign
(China Daily)
HONG KONG: Chief Executive Tung Chee-hwa announced yesterday that he has officially submitted his resignation to Beijing citing health problems.
At a press conference in Hong Kong, Tung, two years into his second term in office, said he had tendered his resignation just one hour before making his decision public.
Hong Kong Chief Executive Tung chee-hwa gestures during a press conference at the Central Government Offices (CGO) building in Hong Kong, 10 March 2005 announcing his resignation. Hong Kong's leader Tung Chee-hwa submitted his resignation March 10, citing health reasons for stepping down early. [AFP]
"My health was obviously not as good as it used to be because of long- time hard work," said Tung.
Chief Secretary for Administration, Donald Tsang, will take over as acting chief executive until a new appointment is made later in the year.
Tung will be in Beijing Friday to attend the third session of the 10th National Committee of the Chinese People's Political Consultative Conference (CPPCC) until it wraps up tomorrow.
At the news conference, Tung stressed that for the overall interests of the Hong Kong Special Administrative Region and the nation, he had decided to submit his resignation .
He expects the central government to consider and accept his request.
President Hu Jintao (R) meets with Tung Chee Hwa in Beijing March 5, 2005.[Xinhua]
Tung stressed that it had been a lifetime pleasure for him to render service for the nation and Hong Kong in the post of chief executive.
He thanked Hong Kong's citizens for the opportunities they had given and central government for the trust they placed in him during his term of office.
Tung, 67, was elected as the first chief executive of the Hong Kong Special Administrative Region in December 1996, and re-elected for a second term in February 2002.
He was inducted as a member of the 10th National Committee of the CPPCC on February 28, 2005.
Today's Top News Top China News
Farming sets goals to raise productivity
Steps to narrow rich-poor gap needed
Peninsula likely to top Rice's first visit
Money talks in capturing drug suspect
Suicide bomber kills 47 at Iraq funeral
First private airline takes off today
Stepping up exchanges across Straits
Reform changes farmers' lives
Migrant workers miss out on cultural experiences
Go to Another Section
select hot link News Center China World Business Life Sports Most Popular Photo Gallery
Leaders meet with Tung Chee Hwa
Tung highlights growth, harmony in address
Tung's team of advisers gathers in Guangzhou
HK's Tung desires good LegCo ties
Tung to HK protestors: Your voices heard
Forecast on Tung 'exit' dismissed
It is time to prepare for Beijing - 2008


Mr Rossiter is a recurring character who appears occasionally within the show. He is the landlord to Stephen, Madison and Ella and owns the Apartment building which houses their apartments. He is shown to be angry and despise Basil and the others, often just turning up to demand the rent.
In Ella he is showing people around the building who are interested in renting Ella's Flat, before it was hers, but all of the people kept fleeing when they saw Basil, who they mistook for diffrent animals. Getting annoyed with this happening over and over again, he shut Basil and Steven in the bathroom of their flat but Basil and Stephen escaped through the air conditioning system via a "handily-placed grill" on the ceiling.
In another episode the gang have turned the flat into a popular hotel by adversing the flat as "The Worst Hotel in the World" where the guests stay in cramped room and the staff, the gang, are rude and insulting to them. This is shut down at the end of the episode by Mr Rossiter when he turns up in a devil costume and charges them £100 for every customer whose in the flat.
Also in one of the others, the boys attacked him, thinking he was a thief. He returns in the twelfth episode of Series 6.He is loosely based on Leonard Rossiter who appeared in Rising Damp as a landlord.


Installing a driveway gate at your home not only adds a welcoming or elegant entrance, but it can also add security as well. There are many different types of driveway gates on the market and one can be found that will suit just about any home or driveway.
Depending upon the type of driveway you install, you can also add a bit of value to your home if you decide to sell. Potential home buyers view driveway gates as a nice upgrade and make them feel better about the safety of the driveway and home.
In this Driveway Guides article, we will discuss the most popular driveway gates as well as how much you can expect to pay to have a driveway contractor install one at your home.
Keep in mind while looking at these prices that installation costs will vary by area. You can save upwards of $500 if you choose to install the posts yourself when you consider this is the most labor intensive part of the installation process.
Some people think that an aluminum gate is the "cheap" way out, but they have come a long way and are now available in a wide variety of strengths. They can also be found in many different types such as arched, tubular and slide and be a good choice if you are on a budget. Since there are many different types and sizes, costs will vary quite a bit.
Wood or cedar driveway gates have become very popular in the last few years and they will fit perfectly with many different types of fences. You do not need to have a split-rail fence to install a wood gate because they fit well with wrought iron or metal as well. Wood gates are not only strong and sturdy, but also add curb appeal to your property. There are quite a variety of wood gates from simple split-rail types, plank and also large, ornate carved wood.


text:
<?xml version="1.0" encoding="iso-8859-1"?>
<!DOCTYPE html
     PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
     "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

  <title><=> (Time)</title>
  <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
  <link rel="stylesheet" href="../.././rdoc-style.css" type="text/css" media="screen" />

      # File lib/rake.rb, line 1647
1647:   def <=>(other)
1648:     if Rake::EarlyTime === other
1649:       - other.<=>(self)
1650:     else
1651:       rake_original_time_compare(other)
1652:     end
1653:   end

redpajama_set_name: RedPajamaGithub


text:
By Jamie Stewart
Local dancer shines on national stage
Jamie Stewart | The Shafter Press
Aliyah Ramos Esquivel will be traveling the United States in the coming months.
Aliyah Ramos Esquivel, a dancer who lives in Shafter, was recognized recently with a prestigious Team Scholarship award from the Hollywood Vibe Company. Hollywood Vibe is a national dance convention/competition company that tours the United States with some of the top industry choreographers and dancers in the nation.
Esquivel's scholarship enabled her to audition for the company nationals in July. She was chosen to be a part of the company's 23-member cast.
She will now be making make appearances all over the country, opening and closing shows, assisting choreographers, taking class, and will have opportunities to enhance her dance career. She is the only dancer in Kern County represented on the tour.
Esquivel with the creative director of Hollywood Vibe, Liz Imperio.
Esquival's home dance studio is Precision Dance Center in Delano. She has been dancing for eight years. She began dancing because her parents wanted her to be involved in something.
Her teacher, Jill Jimenez, has been with Esquivel from the beginning. "I saw the potential in Aliyah very early on," said Jimenez. Esquivel was soon involved in competitive dancing and won numerous awards as a soloist and a group member.
She has received several recognitions at prestigious dance conventions such as Velocity and Radix.
Esquivel will be traveling for the next several months, performing at conventions and shows in cities such as Scottsdale, Arizona, in October; Layton, Utah, in November; Baton Rouge, Louisiana, in January; and Dallas in February.


text:
Mediterranean flavors coming soon to Hesburgh Center for International Studies
When it comes to food options on campus, faculty, staff and students share the opinion that offerings should be flavorful and healthy.
Mission accomplished in the Duncan Student Center, but what's on the horizon?
Building on the popularity of the Modern Market, Star Ginger and Hagerty Family Cafe micro-restaurants that opened in Duncan in January, Campus Dining will officially welcome Garbanzo Mediterranean Fresh to the Hesburgh Center for International Studies at the beginning of the spring semester.
The nation's leading Mediterranean cuisine brand, Garbanzo is widely known for its commitment to sustainability as well as its scratch-made, preservative-free menu options including pitas, gyros, salads, flatbreads and more.
"We're excited to bring a new dining option to campus that's delicious, healthy and decidedly different," said Chris Abayasinghe, senior director of campus dining. "Garbanzo's recipes are crafted using fresh, authentic ingredients, which is why every dish is so flavorful. In addition to creating more variety on campus, it will be perfect for students, faculty and staff who are looking for a quick, wholesome, healthy bite to eat either on-the-go or while relaxing in the Hesburgh Center."
R. Scott Appleby, professor of history and the Marilyn Keough Dean of the Keough School of Global Affairs, says Garbanzo will be a welcome addition to the south side of campus. "We are just delighted to partner with Campus Dining on this project and to have Garbanzo next door to Keough (Jenkins Nanovic Halls). The menu looks great and we fully intend to take advantage of all the fresh, healthy Mediterranean cuisine this operation is bringing to the campus community."
The restaurant is slated to open in the spring, filling the space now occupied by Au Bon Pain Express.
Here are some Garbanzo menu options to look forward to:
Freshly baked pitas
This staple of Mediterranean cuisine will be freshly baked at Garbanzo throughout the day. Choose one as an entree and fill with fresh, nutrient-packed ingredients or add it as a side to complement your salad or plate.
Top-quality proteins
With protein options for carnivores and vegetarians alike, choose from grilled chicken or steak, gyro meat, falafel, chicken or steak kabobs and portobello mushrooms.
Veggies and sauce
Adorn your pita, salad, plate or flatbread with a whole host of nutrient-packed vegetable and sauce options featuring freshly prepared Mediterranean staples like hummus, baba ganoush, tzatziki sauce and feta cheese.
Originally published by Cory Hankins at dining.nd.edu on August 22, 2018.
Author: Cory Hankins
Posted In: Staff Stories


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WINNER: Christopher Wood competing in the PGA Pro-Am Series on the Sunshine Coast at Headland Golf Club today. PGA Media
Wood rises to take out Headland Pro-Am
11th Jul 2019 12:01 AM
GOLF: Despite a shaky start on the second hole, Christopher Wood quickly bounced back strong to take out the Headland Pro-Am yesterday.
Wood shot six under par, to triumph in a "classy" field that included European Tour winner Marcus Fraser, in-from player Matthew Millar and local hopes Charlie Dann and Jacob Boyce.
Dann, and Braden Becker finished -5 for second.
The win earned him $11,000 in prizemoney, with a chance for more with play starting today at Maroochy River, the second tournament on the Sunshine Coast.
"It was a good day out there, I bogeyed my second hole but apart from that I didn't really do much wrong," Wood said.
"None of us like making bogeys, but I knew it was early, and had 16 holes ahead of me. So I just stayed patient.
"I was able to chip it in for an eagle on the 10th, and then hit straight birdies on 11, 12 and 13, that really set up the round."
The Headland tournament is one of only a handful Wood has competed in this Pro-Am series, as the 28-year-old predominantly competes in China.
He said the field held up well considering all the recent rain.
"It was a bit wet, but very playable and the greens rolled out nicely," he said.
"The fairways were quite nice, overall it was really enjoyable.
"I'm looking forward to playing at Maroochy River too, but I seem to find it challenging. It's a windy course, so you have to really control the flight of the ball and putt and chip well. Wind is a big factor."
christopher wood golf headland golf club headland pro-am maroochy river golf club pga pro-am series sunshine coast


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The Fifth Estate: Disney/ABC Media Empire David And Goliath Demonization Of Assange
It would seem unusual to label a dramatic feature a biopic, when the content is based solely on the accusations of an admitted adversary with a vendetta against the subject in question. But once again, The Fifth Estate, an unabashed demonization of the sinister state secrets whistleblowing website Wikileaks and its head hacker extraordinaire Julian Assange, is all about what Hollywood does best - the one side to every story school of moviemaking. Or is it?
Features of Arts Express: Expression In The Arts are hosted by Prairie Miller, and air nationally on the Pacifica National Radio Network and WBAI/Affiliate Stations, including WPRR Public Reality Radio. And if you'd like to Express yourself too, you can write to: ArtsExpressradio@gmail.com
Mediastan Movie Review: Ferreting Out The Presstitutes
Sometimes it can be said that a documentary is exemplary for not accomplishing what is set out to do. And the Julian Assange Wikileaks production Mediastan may have succeeded in not doing just that. As those Wikileaks foot soldiers fail in their mission to find remote and presumably uncorrupted media organizations courageous enough to publish the damning cables leaked by far more courageous whistleblowers than these press outlets prove to be.
Increasingly doomed to extinction on the endangered list, so to speak, is mainstream journalism. The victim of a tug of war between the corporations gobbling up the pitiful remains, and incestuously connected powerful political interests - too often one and the same - journalism has given rise in the wake of this fate to its deplorable mutation - presstitutes.
Mother Of George: Exquisitely Drawn Film Portrait Of Female Identity Theft Disappeared By Marriage
Simply telling the story of Andrew Dosunmus' enticing feature Mother of George, doesn't convey the movie's extraordinary visual power. The film's poetic impact begins with preparations for a colorful Yoruba wedding in Brooklyn, with close-ups of the wedding parties of Ike (Danai Gurira), the newly arrived Nigerian bride, and Ayo (Isaach de Bankole), her groom.
Ayo works with his younger brother Biyi (Tony Okunghowa) at a restaurant overseen by their mother, (Bukky Ajayi). After the ceremonies, the women gather around the bride, giving her child rearing tips. While the men counsel the groom on how to hide his infidelities.
Thus, director Nigerian Dosunmu and screenwriter Daniel Picoult carefully begin to document the friction that ensues when the rigid gender expectations of Nigerian tradition clash with more liberal opportunities that Ike's new home offers. And when after eighteen months, Ike hasn't become pregnant, Ayo becomes enraged when she offers to get a job to pay for a fertility specialist. He refuses to even go to the doctor, fearing it will reveal that he's infertile.
By this time, Ma Ayo (Bukky Ajay), who holds on to old fashioned ideas, believes her own happiness lies in having a grandchild. Even if it means that Ike must provide one with another partner. Thus she enlists Ayo's big brother Biyi - who has been keeping secret his affair with Ayo's best friend Sade (Yaya Alafia), fearing that her Western values will offend his family.
One of the film's greatest achievements it to present exotic characters with a familial dilemma that crosses international borders and cultures. A further delight of this drama, is the way filmmaker Dosunmu and Bradford Young's gorgeous cinematography highlight Ike in colorful Nigerian dresses. Which make her seem like some distant goddess on congested Brooklyn streets.
Ike is also shot in close-up, so that we see the world from her narrow perspective. As she stands out from the crowd, like a marvelous exotic addition to Brooklyn's landscape.
Jan Aaron writes for Education Update, and is a member of the Women Film Critics Circle.
The Fifth Estate: Disney/ABC Media Empire David An...
Mediastan Movie Review: Ferreting Out The Presstit...
Mother Of George: Exquisitely Drawn Film Portrait ...