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text:
Q: Populate SQL table column from another table I have a jqGrid(latest free jqGrid version) table filled with data from the database(MS SQL) using a Java REST service. One of the jqGrid columns its a drop down list with 6 options.
My asignment is to create another database table that contains the Drop Down List values and using foreign keys / primary keys i have to automaticly populate the first table DDL values.
I am failing to understand the logic behind it. Can someone expain this to me? How can i achieve this. Do i only send a ID from jqGrid and depending on that ID(1,2,..,6) it chooses what to set in table#1 DDL column(comparing the id sent with the ID's of the table that contains the DDL Values)?
I am getting the feeling i am not expressing myself well... Hope you guys understand me.

A: We can start with the database table. It could look like
CREATE TABLE dbo.OrderStatus (
    Id int IDENTITY NOT NULL,
    Name nvarchar(100) NOT NULL,
    CONSTRAINT PK_LT_OrderStatus PRIMARY KEY CLUSTERED (Id),
    CONSTRAINT UC_LT_OrderStatus_Name UNIQUE NONCLUSTERED (Name)
)

It allows to address any Item of such OrderStatus table by Id or by Name. The UNIQUE CONSTRAINT don't permit to add name duplicates. Another table Order can have column
CREATE TABLE dbo.Order (
    Id int IDENTITY NOT NULL,
    OrderStatusId int NOT NULL,
    ...
)

ALTER TABLE dbo.Order  WITH CHECK ADD  CONSTRAINT FK_Order_OrderStatus
FOREIGN KEY(OrderStatusId) REFERENCES dbo.OrderStatus (Id)

During filling the grid with the data you have two main options: using OrderStatusId in the data or the usage of the corresponding Name from dbo.OrderStatus:
SELECT Id,OrderStatusId, ...  FROM dbo.Order

or
SELECT Id,os.Name AS OrderStatus, ...
FROM dbo.Order AS o
    INNER JOIN dbo.OrderStatus AS os ON os.Id=o.OrderStatusId

If you decide to fill the grid with ids (OrderStatusId values) then you will have to use formatter: "select" to display the text in the corresponding column (see here). It required that you would have to have editoptions.value filled with all different values from dbo.OrderStatus. The best way to implement this would be to extend the response from the server for filling the grid with your custom data and to use beforeProcessing to set editoptions.value. I described the scenario in the answer. I'll remind you it below.
Let us the response from the server looks like
{
    "rows": [{...}, {...}]
}

If the returned data looks just like
[{...}, {...}]

then you should include the wrapping. I suggest that you made
SELECT Id,Name FROM dbo.OrderStatus

additionally to making the main select from dbo.Order (SELECT * FROM dbo.Order) and you place both results in the server response:
{
    "orderStatus": [{"id":1, "name":"Pending"}, ...],
    "rows": [{...}, {...}]
}

To process orderStatus you need to add the following beforeProcessing, which read orderStatus and set editoptions.value of the orderStatus column of the grid:
beforeProcessing: function (response) {
    var $self = $(this), orderStatus = response.orderStatus, i, values = "";
    if (orderStatus != null && orderStatus.length > 0) {
        for (i = 0; i < orderStatus.length; i++) {
            if (values.length > 0) {
                values += ";";
            }
            values += orderStatus[i].id + ":" + orderStatus[i].name;
        }
        $self.jqGrid("setColProp", "orderStatus", {
            editoptions {
                value: values
            }
        });
        if (this.ftoolbar) { // filter toolbar exist
            $self.jqGrid("destroyFilterToolbar");
            $self.jqGrid("filterToolbar");
        }
    }
}

The above code is not tested, but I hope that the main idea should be clear from it.

redpajama_set_name: RedPajamaStackExchange


The 1500-seat theater is located on top of parking building with its lobby located on the 4th floor of Esplanade shopping center. It is planned and designed to serve as an international-standard musical theater. The design of the theater creates the uniqueness for the theater by connecting to its context, Thailand and by the use of traditional Thai element in a different approach with colors and patterns that express luxurious and prestige Thai traditional arts.


The plot of the series "Sick Note" will turn around – a young man with a pathological inclination to lie. When Daniel is diagnosed with esophageal cancer, everyone begins to take part and care, trying to alleviate his suffering. But when it turns out that the diagnosis was false, Daniel is not going to give up his status of "dying." And to support this lie, he attracts to the collaboration of his oncologist Dr. Glennis – the most incompetent doctor in the world.


text:
Shooter fires at downtown Royal Oak hookah lounge
One employee of Hookah Joe's was shot in the head but the injury was minor, police say.
Shooter fires at downtown Royal Oak hookah lounge One employee of Hookah Joe's was shot in the head but the injury was minor, police say. Check out this story on Freep.com: https://on.freep.com/1pDk50C
Matt Helms, Detroit Free Press Published 5:56 p.m. ET March 20, 2016 | Updated 7:06 p.m. ET March 20, 2016
Close-up of emergency lights(Photo: Getty Images)
Royal Oak police are looking for two men who they say shot several rounds into a hookah lounge downtown early Sunday.
The shooting around 3:30 a.m. at Hookah Joe's, 734 S. Washington, happened after an argument inside the lounge, police said in a news release. The suspects fired one round at an individual outside the facility and then pulled their vehicle to the front of the shop and fired three or four more rounds.
One employee inside was hit in the head but the injury was minor, police said.
According to police, there was an altercation between the two suspects and another patron inside Hookah Joe's just before the shooting. One of the suspects pulled out a handgun, and the other motioned as if he had one in his waistband, police said.
Security personnel escorted the suspects out of the lounge, and they were confronted by the other patron as they walked to their vehicle parked on South Washington, police said. One of the suspects fired a single round at that patron, and as he re-entered the front of the business, the suspects pulled up in front and fired three or four more rounds.
The employee who was shot was not involved in the altercation, police said. While there were numerous customers inside, no other injuries were reported.
Police said witnesses reported that the suspects fled in a silver or gray vehicle, likely a Dodge Charger, with chrome rims, tinted windows and a dent on the passenger-side door.
One suspect is described as 6 feet or 6 feet 3 with a thin build; the other is described as 6 feet with a medium build and long hair in a ponytail.
Anyone with information is asked to call Royal Oak police at 248-246-3500.
Sheriff: Uber lawsuit in Kalamazoo shooting is a hoax
Contact Matt Helms: mhelms@freepress.com or on Twitter: @matthelms
Read or Share this story: https://on.freep.com/1pDk50C
Why huge Michigan voter turnout could turn into national embarrassment
Wife's divorce filing reveals details of chemist, father now accused of cannibalism
Warren resident, 27, killed in 4-car crash in Troy
Poll: Mike Bloomberg holds 7-point lead on Trump in Michigan

redpajama_set_name: RedPajamaCommonCrawl


I can't believe I am writing about my trip to Paris as I am watching the news about Notre-Dame being on fire! It is so sad! Notre- Dame is always a must see landmark that all visitors go to pay their respects. It survived 8 centuries, and to see it just burn is very sad.
One of my favorite parks in Paris is the Jardin de Luxembourg. It's huge and beautiful! Filled with flowers and trees, you forget that you are in a city. Very much the same feeling as being in Central Park, but a la French. What makes it my favorites, is that there is always a photo exhibit showcased along the gate of the park. One of my life goals is to have one of my photos hung in one of their exhibits.
This was a mother daughter trip. Last mother/daughter trip we took was Paris too, back in 2017. My mom lovessss PARIS, I think she might have been Parisian in another life. It really is a beautiful city. The weather was a bit crisp, but tolerable when dressed properly. My mom was my muse while on the trip but also artistically inspiring. We were walking by a group of empty chairs and she mentioned how she always liked to photograph empty chairs, it was like they were having a dialogue. So of course, I am now getting obsessed with photographing chairs. Like mother, like daughter.
Another subject or theme I was focused on during the trip was architecture and the color of walls or doors. I love the contrast of bright colors with muted colors. A lot of the time it was gray and gloomy, so I liked finding brightness in the darkness.
Spring was in full bloom and there were little flower gardens you could find while walking the Parisian streets. I usually don't take many photos of flowers and leaves, except for in the fall. I must admit they were beautiful to photograph!
But I do agree with my mom, Paris is a beautiful city. And as I continue to see Notre-Dame in flames on the news, I cannot stress enough how important it is to travel and learn about history all over the world. You never know when they will vanish.


text:
The new dELiA*s catalog has a total 'zine vibe
By Sammy nickalls
For us '90s kids, we've had a lot of up-and-downs when it comes to our fave store, dELiA*s. Around this time last year, we discovered that it had declared bankruptcy and liquidated all its merchandise. And we cried sad tears. But then, we found out this April that it's returning as an e-shop! And we cried happy tears! And THEN, news broke in August that we can get the catalog delivered monthly to our door, just like the old days, and GUESS WHAT WE DID. (Spoiler: It involves tears.)
Now, we've got a taste of the new catalogue, and it's better than we could have ever imagined. It's going to have a 'zine feel, with editorial spreads and content, like the Rookie of our '90s dreams. For the first edition of the revamped 'zine-catalog, the subject will be 16-year-old Flynn McGarry, who released a cookbook when he was just 11 and went on to own his own NYC restaurant.
Although all of us are super nostalgic for our dELiA*s days, we are no longer the brand's focus — which is exactly why they're revamping their image and style. "dELiA*s isn't relying on nostalgic millennials anymore," dELiA*s executive vice president and chief merchandising officer, Patricia Johnson, said in a press release, according to Racked. "We're sort of enhancing what dELiA*s was, so we're focusing in on things we think are important to [the customer], like food."
The catalog will be focusing on teens like McGarry to be a role model for the target dELiA*s customer. "For the past few months, we've been reaching out to collaborate with young talent that our girls can aspire to," Johnson said in the release.
Well, one thing's for sure: No matter their target customer, many nostalgic millennials (myself included) will be signing up for the new catalog to check it out. To get it delivered to you every month, you can sign up here. Be still, our beating '90s hearts.
Our beloved dELiA*s is back and we couldn't be happier about it
(Image via Twitter.)
90's • nostalgia • shopping


GUIDE: Where Should Christians Get Their News?
To avoid the omissions of the mainstream media, we should look at the publications of advocacy and watchdog groups that bring to our attention worthy items that the mainstream media does not cover. This includes ethnic-oriented groups (e.g. the Anti-Defamation League), human rights groups such as Amnesty Intl, and environmental and animal rights groups.
Some people will scoff at this, but the first 15 minutes of this show is a daily wrap-up of the news from a Christian perspective. No RSS.
This compact news weekly is inexpensive and covers world news with a balanced perspective. Minimal online content, though.
Nice Christian view of what is important, and how evangelicals view world events. Too bad their online blog doesn't have an RSS feed! BLINO – BLog In Name Only.
This blog really has a nice stream of news from a Christian perspective. RSS Feed What are your favorite news sources?
I never thought I would see NPR and the 700 Club together on anyone's list of anything!Most conservatives (with cable/dish) get their news from FOX News. I don't have either and I don't get home in time to watch the evening news (not that I would if I could).I tend to get most of my news from blogs, online news services, etc. I go to a variety of blogs with different slants, that way I know what I am getting.
NPR has an unfair reputation among conservatives. To their own hurt, they miss a lot of great reporting on npr.I don't watch fox news because its coverage is so overtly biased as to be annoying, kind of like Bill O'Reilly. I'm sure some of it is fine, but I don't care for it much. Maybe I'm missing something, but it seems to be the conservative equivalent of Dan Rather's reporting. Certainly not balanced.


Best Match! Charles H. Fitzpatrick, OD is a Optometry specialist in Cherry Hill, NJ. This doctor matches your preferences for Specialty.
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Family-owned Australian battery specialist Master Instruments has completed its move to new larger premises in South Western Sydney.
The move from Marrickville to the new headquarters, in Sheridan Close, Milperra, gives the 75-year-old company ample space to build on its growing market share.
Master Instruments manufactures, imports and distributes Australia's largest range of industrial and consumer primary and secondary batteries and battery packs. The new 5500m2 building houses battery manufacturing lines and packaging, engineering and tool shop, electronics laboratories, R&D, meter production and two warehouses, in addition to administration and sales.


text:
\section*{I. Introduction}

\hskip 20pt
Recently, the theories of quantum enveloping algebras and quantum
groups have drawn wide concerns, and made great progress $^{[1,2]}$.
For a generic $q$, there have been standard methods for calculating
the highest weight representations and Clebsch-Gordan coefficients
of quantum enveloping algebras $^{[3,4]}$. Based on them, a series
of solutions $R_{q}$ of the simple Yang-Baxter equation (without
spectral parameter) and solutions $R_{q}(x)$ of the Yang-Baxter
equation $^{[5]}$ were found. In terms of the quantum double, the
general form of universal ${\cal R}$ matrix were obtained in principle
$^{[3]}$. For $q$ being a root of unity, the representation theory was
studied, too $^{[6]}$. The representation of quantum affine
algebras and their applications to the solvable lattice models
has been made great advance $^{[7]}$. On the other hand, following
the general idea of Connes on the non-commutative geometry $^{[8]}$,
Woronowicz $^{[9]}$ elaborated the framework of the non-commutative
differential calculus. He introduced the bimodule over the quantum
groups and presented varies theorems concerning the differential
forms and exterior derivative. The $q$-deformed gauge theory was
studied from several different viewpoints $^{[8-10]}$.

A quantum group is introduced as a Hopf algebra ${\cal A}=Fun_{q}(G)$
which is both non-commutative and non-cocommutative. It is
the continuous deformation of the Hopf algebra of the functions
on a Lie group, and is freely generated by the non-commutative
matrix elements $T^{a}_{~b}$:
$$(R_{q})^{ca}_{~rs}T^{r}_{~b}T^{s}_{~d}~=~
T^{a}_{~r}T^{c}_{~s}(R_{q})^{sr}_{~bd} \eqno (1.1) $$

\noindent
where the matrix $T$ takes value in the minimal representation, and
$R_{q}$ is the solution of the simple Yang-Baxter equation corresponding
to the minimal representation.

Faddeev-Reshetikhin-Takhtajan $^{[11]}$ introduced two linear functionals
$L^{\pm}$ which belong to the dual Hopf algebra ${\cal A'}$. They are
defined by their values on the elements $T^{a}_{~b}$:

$$\left(L^{+}\right)^{a}_{~b}\left(T^{c}_{~d}\right)~=~
\left(R_{q}^{-1}\right)^{ac}_{~bd},~~~~~
\left(L^{-}\right)^{a}_{~b}\left(T^{c}_{~d}\right)~=~
\left(R_{q}\right)^{ca}_{~db} \eqno (1.2) $$

As is well known, the quantum enveloping algebra is also a
Hopf algebra dual to ${\cal A}$, and the eq.(1.1) implies, in fact,
this dual relation (see Sec.II), so it must tightly connect with
the operators $L^{\pm}$. Ref.[11] gave their relations for
$U_{q}A_{1}$, but their generalization is not trivial. Some
authors tried to make clear the profound relations between
quantum groups and quantum enveloping algebras $^{[12]}$.
In this paper, after a further discussion on those relations
we carefully study the concrete expressions of the operators $L^{\pm}$
by the generators in a quantum enveloping algebra, and explicitly
give the expressions in two typical examples:
$U_{q}A_{N}$ and $U_{q}G_{2}$ algebras.

\vspace{10mm}
\section*{II. The dual relation}

\hskip 20pt
The generators $t_{j}$, $e_{j}$ and $f_{j}$ of a quantum enveloping
algebra $U_{q}{\cal G}$ satisfy the following algebraic relations:
$$\begin{array}{l}
t_{i}t_{j}~=~t_{j}t_{i},~~~~t_{j}^{-1}t_{j}~=~t_{j}t_{j}^{-1}~=~
{\bf 1},\\
t_{i}e_{j}~=~q_{i}^{a_{ij}}e_{j}t_{i},~~~~
t_{i}f_{j}~=~q_{i}^{-a_{ij}}f_{j}t_{i},~~~~q_{i}~=~q^{d_{i}}\\
{}~[e_{i},f_{j}]~=~\delta_{ij}\omega_{j}^{-1}\left(
t_{j}-t_{j}^{-1}\right),~~~~~~~~\omega_{j}~=~q_{i}-q_{i}^{-1}\\
\displaystyle \sum_{n=0}^{1-a_{ij}}~(-1)^{n}\left[\begin{array}{c}
1-a_{ij}\\[-2mm] n\end{array} \right]_{q_{i}}~e^{1-a_{ij}-n}
e_{j}e_{i}^{n}~=~0,~~~~i\neq j \\
\displaystyle \sum_{n=0}^{1-a_{ij}}~(-1)^{n}\left[\begin{array}{c}
1-a_{ij}\\[-2mm] n\end{array} \right]_{q_{i}}~f^{1-a_{ij}-n}
f_{j}f_{i}^{n}~=~0,~~~~i\neq j
\end{array} \eqno (2.1) $$

\noindent
where $a_{ij}$ is the Cartan matrix element of a Lie algebra
${\cal G}$, and $d_{j}$ is half-length of the simple root. For
the longer simple root, $d_{j}=1$. As a Hopf algebra, the coproduct
$\Delta$, the counit $\varepsilon$ and the antipode $S$ of the
generators are defined as follows:
$$\begin{array}{c}
\Delta(t_{j})~=~t_{j}\otimes t_{j},~~~~
\Delta(e_{j})~=~e_{j}\otimes {\bf 1}+t_{j}\otimes e_{j}, \\
\Delta(f_{j})~=~f_{j}\otimes t_{j}^{-1}+{\bf 1}\otimes f_{j}, \\
\varepsilon(t_{j})~=~1,~~~~\varepsilon(e_{j})~=~\varepsilon(f_{j})~=~0,\\
S(t_{j})~=~t_{j}^{-1},~~~~S(e_{j})~=~-t_{j}^{-1}e_{j},~~~~
S(f_{j})~=~-f_{j}t_{j},
\end{array} \eqno (2.2) $$

In fact, the representation matrix elements
$D^{[\lambda]}_{q}(\alpha)_{\mu \nu}$ of a quantum enveloping
algebra give the dual relation between the generators
$\alpha\in U_{q}{\cal G}$ and the non-commutative quantity
$\left(T^{[\lambda]}\right)^{\mu}_{~\nu}$ of the quantum group
in the corresponding representation:
$$\begin{array}{c}
\left\langle \alpha, \left(T^{[\lambda]}\right)^{\mu}_{~\nu}
\right\rangle~=~D^{[\lambda]}_{q}(\alpha)_{\mu \nu},~~~~
\alpha \in U_{q}{\cal G}
\end{array} \eqno (2.3) $$

\noindent
When $[\lambda]$ is the minimal representation,
$\left(T^{[\lambda]}\right)^{\mu}_{~\nu}$ becomes $T^{a}_{~b}$.

The universal ${\cal R}$ matrix can be expressed as
$${\cal R}~=~\displaystyle \sum_{i}~u^{i}\otimes v_{i}~\in~
U_{q}{\cal G}\otimes U_{q}{\cal G} \eqno (2.4) $$

\noindent
where $u^{i}$ and $v_{i}$ are the dual bases of up- and down-Borel
subalgebras, respectively. Take the value of ${\cal R}$ matrix in
the minimal representation, we get the $R_{q}$  matrix:
$$\langle {\cal R}, T^{c}_{~d}\otimes T^{a}_{~b}\rangle
{}~=~\left(R_{q}\right)^{ca}_{~db} \eqno (2.5) $$

The universal ${\cal R}$ matrix satisfies $^{[3,4]}$:
$${\cal R}\Delta(\alpha)~=~\left(P\circ \Delta(\alpha) \right){\cal R},
{}~~~~\alpha\in U_{q}{\cal G} \eqno (2.6) $$
$${\cal R}^{-1}~=~\left({\rm id}\otimes S^{-1}\right){\cal R}
\eqno (2.7) $$

\noindent
where  $P$ is the space transposition operator. Substituting eq.(2.6)
into eq.(2.5), we get
$$\begin{array}{l}
R^{ca}_{~rs}~\langle \Delta(\alpha), T^{r}_{~b}\otimes T^{s}_{~d}\rangle
{}~=~\langle {\cal R}\Delta(\alpha), T^{c}_{~b}\otimes T^{a}_{~d}\rangle \\
{}~=~\left\langle \left(P\circ \Delta(\alpha)\right){\cal R},
T^{c}_{~b}\otimes T^{a}_{~d}\right\rangle
{}~=~\langle \Delta(\alpha), T^{a}_{~s}\otimes T^{c}_{~r}\rangle
R^{rs}_{~bd} \end{array} \eqno (2.8) $$

\noindent
Due to arbitrariness of $\alpha$, one can immediately get eq.(1.1)
from eq.(2.8). Similarly, it is easy to show that equation (1.1)
holds for any other representation. So the consistent condition
(1.1) is just the direct reflection of the dual relation between
a quantum group and a quantum enveloping algebra.

Now we rewrite eq.(1.2) as the form of dual relation:
$$\left\langle \left(L^{-}\right)^{a}_{~b}, T^{c}_{~d}\right\rangle
{}~=~\left(R_{q}\right)^{ca}_{~db} $$

\noindent
Comparing it with eq.(2.5), we have
$$\left(L^{-}\right)^{a}_{~b}~=~
\langle {\cal R}, {\rm id} \otimes T^{a}_{~b}\rangle \eqno (2.9)$$

\noindent
Similarly,
$$\left(L^{+}\right)^{a}_{~b}~=~
\langle {\cal R}^{-1}, T^{a}_{~b}\otimes {\rm id}\rangle $$

\noindent
{}From eq.(2.7) we have:
$$\left(SL^{+}\right)^{a}_{~b}~=~
\langle {\cal R}, T^{a}_{~b}\otimes {\rm id}\rangle \eqno (2.10)$$

\noindent
In a different notation, the analogous relations were also given in
Ref.[13]. By making use of the general form of
the universal ${\cal R}$ matrix given by Jimbo $^{[3]}$, one can calculate
the explicit relations between $\left(L^{\pm}\right)^{a}_{~b}$ and
the generators in a quantum enveloping algebra from eqs.(2.9) and (2.10).

Denote by $Q_{+}$ the set of all the non-negative integral combinations
of simple roots,
$$Q_{+}~=~\left\{\beta|\beta=\displaystyle \sum_{j} m_{j} {\bf r}_{j},
m_{j}\in {\bar {\bf Z}_{-}} \right\} $$

\noindent
For a given $\beta \in Q_{+}$, we define two finite-dimensional
spaces $N_{\beta}^{\pm}$:
$$\begin{array}{ll}
N_{\beta}^{+}:~~~~&\left\{e_{i_{1}}e_{i_{2}}\cdots e_{i_{m}}|
\displaystyle \sum_{j}{\bf r}_{i_{j}}=\beta\right\}\\
N_{\beta}^{-}:~~~~&\left\{f_{i_{1}}f_{i_{2}}\cdots f_{i_{m}}|
\displaystyle \sum_{j}{\bf r}_{i_{j}}=\beta\right\}
\end{array} $$

\noindent
where ${\bf r}_{i_{j}}$ is the simple root corresponding to the
generators $e_{i_{j}}$ and $f_{i_{j}}$. Those two spaces both
are Hopf algebras, and they are associated and dual to each
other $^{[4]}$. In these dual spaces we can adopt the following dual
bases:
$$u^{i}\in N_{\beta}^{+},~~~~v_{i}\in N_{\beta}^{-},$$

\noindent
so that the general form of universal ${\cal R}$ matrix can be formally
 expressed as $^{[2,4]}$:
$${\cal R}~=~\left(\displaystyle \sum_{\beta \in Q_{+}}~L_{\beta} \right)
{}~q^{-H},~~~~
L_{\beta}~=~\displaystyle \sum_{i}~u^{i}\otimes v_{i}.
\eqno (2.11) $$

\noindent
A representation is called integrable if:

i) $V=\displaystyle \oplus_{\mu} V_{\mu}$
$$V_{\mu}~=~\left\{v_{\mu}\in V|t_{j}v_{\mu}=q^{\mu_{j}}_{j} v_{\mu},
{}~~\mu =\sum \mu_{j} \lambda_{j},~~1\leq j \leq N\right\}$$

ii) For any $v \in V$, there exists a common $M$ such that:
$$e_{j}^{M}~v~=~0,~~~~f_{j}^{M}~v~=~0. $$

\noindent
And $q^{-H}$ is given by
$$q^{-H} \left(v_{\mu}\otimes v_{\nu}\right)~=~
q^{-\displaystyle \sum_{ij}~d_{i}(a^{-1})_{ij}\mu_{i}\nu_{j}}
\left(v_{\mu}\otimes v_{\nu}\right) \eqno (2.12) $$

\noindent
where $(a^{-1})$ is the inverse Cartan matrix of the Lie
algebra ${\cal G}$.

When calculating $\left(L^{-}\right)^{a}_{~b}$ by eq.(2.9), the action
on the first subspace should keep its operator form, but for the second
subspace, it should be taken value in the minimal representation,
i.e., expressed as matrix form. By making use of this method, we
are able to express $\left(L^{\pm}\right)^{a}_{~b}$
operators by the generators in any quantum enveloping algebra in
principle. In the rest of this paper we will compute the explicit
expressions for two typical quantum enveloping algebras as examples.

\vspace{10mm}
\section*{III. $U_{q}A_{N}$ algebra}

\hskip 20pt
The reason for choosing $U_{q}A_{N}$ algebra as the first example
is that it is the simplest and the most useful qutantum enveloping
algebra. Jimbo $^{[2]}$ pointed out that for $U_{q}A_{N}$ any generator
corresponding to a non-simple root can be expressed as a
linear combination of the generators $e_{j}$ (or $f_{j}$) that
correspond to the simple roots:
$$\begin{array}{l}
E_{j(j+1)}~=~e_{j},~~~~E_{ij}~=~E_{ik} E_{kj}-q E_{kj} E_{ik},\\
E_{(j+1)j}~=~f_{j},~~~~E_{ji}~=~E_{jk} E_{ki}-q^{-1} E_{ki} E_{jk},
 \end{array},~~~~~~i<j<k \eqno (3.1) $$

\noindent
Through calculation, we find that $\left( L^{\pm} \right)^{a}_{~b}$ are
directly related with these generators.

In the minimal representation of $U_{q}A_{N}$, the bases can be
represented by only one index $a$, corresponding to the following
weight:
$$a \longrightarrow {\bf \lambda}_{a}-{\bf \lambda}_{a-1},~~~~
{\bf \lambda}_{0}={\bf \lambda}_{N+1}=0,~~~~1\leq a \leq N+1 \eqno (3.2) $$

\noindent
where ${\bf \lambda}_{a}$ are the fundamental dominant weights of
$U_{q}A_{N}$. The matrix forms of the generators in these bases are
given as follows:
$$\begin{array}{l}
t_{j}~|a\rangle~=~q^{\left(\delta_{aj}-\delta_{(a-1)j}\right)}~|a\rangle\\
e_{j}~|a\rangle~=~|(a+1)\rangle,~~~~
f_{j}~|a\rangle~=~|(a-1)\rangle  \end{array}~~~~1\leq a \leq N+1
 \eqno (3.3) $$

{}From eq.(2.12), we have
$$\begin{array}{l}
q^{-H}~\left(|a\rangle \otimes |b\rangle \right)~=~
\displaystyle \left\{\prod_{j}~t_{j}^{\tau_{jb}}
\otimes {\bf 1} \right\}~\left(|a\rangle \otimes |b\rangle \right) \\
\tau_{jb}~=~(a^{-1})_{j(b-1)}-(a^{-1})_{jb}~=~\left\{\begin{array}{ll}
\displaystyle {j \over N+1}-1,~~~~~&j\geq b \\
\displaystyle {j \over N+1},~~~~~&j< b \end{array} \right.
\end{array} $$

\noindent
or briefly express it as:
$$q^{-H}~\left(|a\rangle \otimes |b\rangle \right)~=~\left\{
\displaystyle \left(\prod_{j=1}^{N}~t_{j}^{j/(N+1)}\right)~
\displaystyle \left(\prod_{k=b}^{N}~t_{k}^{-1} \right)\otimes {\bf 1}
\right\}~\left(|a\rangle \otimes |b\rangle \right) \eqno (3.4) $$

Now we use eq.(2.9) to calculate $\left(L^{-}\right)^{a}_{~b}$. If $a<b$,
$\left(L^{-}\right)^{a}_{~b}$ vanishes. If
$a=b$, $\left(L^{-}\right)^{a}_{~b}$ only contains factor $q^{-H}$
given in eq.(3.4). If $a>b$, there is only one term in the
summation of ${\cal R}$ that has nonvanishing contribution to
$\left(L^{-}\right)^{a}_{~b}$, where $v_{i}$ is:
$$v_{i}~=~f_{a-1}~f_{a-2}~\cdots~f_{b},~~~~a\geq b \eqno (3.5) $$

\noindent
The key to the problem is to find $u^{i}$ dual to the above
$v_{i}$. In the calculation the following dual relations $^{[3,4]}$
are used:
$$\begin{array}{l}
\langle~t_{i}^{n}~,~t_{j}^{m}~\rangle~=~q_{i}^{-a_{ij}},~~~~
\langle~e_{i}~,~f_{j}~\rangle~=~- \delta_{ij}~\omega_{i}^{-1},\\
\langle~e_{i}~,~t_{j}^{m}~\rangle
{}~=~\langle~t_{i}^{n}~,~f_{j}~\rangle~=~0,
\end{array} \eqno (3.6) $$

\noindent
When $(a-b)$ is small, the dual operator $u^{i}$ is easy to
calculate. We can drew the general form of $u^{i}$ from these
$u^{i}$, then prove it by induction. Noting that for $U_{q}A_{N}$
all $\omega_{j}$ are equal to each other and can be denoted by
one symbol $\omega$. The result is:
$$\begin{array}{rl}
u^{i}&=~-\omega q^{a-b-1} \displaystyle \sum_{P}~
(-q)^{-n(P)}e_{p_{1}} e_{p_{2}} \cdots e_{p_{a-b}} \\
&=~(-1)^{a-b} \omega E_{ba} \end{array}~~~~~a>b \eqno (3.7) $$

\noindent
where $P$ is a certain permutation of $(a-b)$ objects, that can
be expressed by a product of some transpositions, and each
transposition moves a smaller number from the right of a bigger
one to the left.
$$P~=~\left(\begin{array}{cccc} (a-1)&(a-2)&\cdots&b\\
p_{1}&p_{2}&\cdots&p_{a-b} \end{array} \right) $$

\noindent
In such a multiplication expression, each transposition appears
at most once. We are only interested in the "neighboring
transposition" that permutes the places of two neighboring
numbers $c$ and $(c-1)$. Two $P$ are called equivalent if
their multiplication expressions contain same neighboring
transpositions without concerning their order.
The summation in eq.(3.6) runs over all inequivalent
permutations include the unit element. $n(P)$ is the
number of neighboring transpositions contained in $P$.

Substituting (3.4), (3.7) into (2.9) and (2.11), we finally get
$$\left(L^{-}\right)^{a}_{~b}~=~\left\{ \begin{array}{ll}
0,&a<b \\
\left\{\displaystyle \prod_{j=1}^{N} t_{j}^{j/(N+1)}\right\}
{}~\left\{\displaystyle \prod_{k=b}^{N} t_{k}^{-1}\right\}, &a=b \\
(-1)^{a-b} \omega E_{ba}
\left\{\displaystyle \prod_{j=1}^{N} t_{j}^{j/(N+1)}\right\}
{}~\left\{\displaystyle \prod_{k=b}^{N} t_{k}^{-1}\right\},~~~~&a>b
\end{array} \right. \eqno (3.8) $$

\noindent
Similarly
$$\left(L^{+}\right)^{a}_{~b}~=~\left\{ \begin{array}{ll}
(-1)^{b-a+1} \omega
\left\{\displaystyle \prod_{j=1}^{N} t_{j}^{-j/(N+1)}\right\}
{}~\left\{\displaystyle \prod_{k=a}^{N} t_{k}\right\} E_{ba},~~~~&a<b\\
\left\{\displaystyle \prod_{j=1}^{N} t_{j}^{-j/(N+1)}\right\}
{}~\left\{\displaystyle \prod_{k=a}^{N} t_{k}\right\}, &a=b \\
0, &a>b \end{array} \right. \eqno (3.9) $$

Note that it is only for $U_{q}A_{N}$ where the generators
corresponding to non-simple roots can be expressed as (3.1), so
that the expressions of $\left(L^{\pm}\right)^{a}_{~b}$
become so simple. For other algebras, the expressions become
much more complicated. $U_{q}G_{2}$ algebra is an example that
will be discussed in the next section.

\vspace{10mm}
\section*{IV. $U_{q}G_{2}$ algebra}

\hskip 20pt
$U_{q}G_{2}$ is another typical example where the lengths of
two simple roots are different. Let
$$\begin{array}{l}
q_{1}~=~q,~~~~q_{2}~=~q^{1/3},~~~~[m]~=~\displaystyle {q_{2}^{m}-q_{2}^{-m}
 \over q_{2}-q_{2}^{-1} },\\
\omega_{2}~=~q_{2}-q_{2}^{-1},~~~~\omega_{1}~=~q_{1}-q_{1}^{-1}
{}~=~[3]\omega_{2} \end{array}
\eqno (4.1) $$

\noindent
The minimal representation of $U_{q}G_{2}$ algebra is given as follows:
$$\begin{array}{l}
D_{q}(t_{1})~=~{\rm diag}\left\{1,~q,~q^{-1},1,~q,~q^{-1},~1\right\},\\
D_{q}(t_{2})~=~{\rm diag}\left\{q_{2},~q_{2}^{-1},~q_{2}^{2},1,~
q_{2}^{-2},~q_{2},~q_{2}^{-1}\right\},\\
D_{q}(e_{1})_{23}~=~D_{q}(f_{1})_{32}=
D_{q}(e_{1})_{56}~=~D_{q}(f_{1})_{65}=1,\\
D_{q}(e_{2})_{12}~=~D_{q}(f_{2})_{21}=
D_{q}(e_{2})_{67}~=~D_{q}(f_{2})_{76}=1,\\
D_{q}(e_{2})_{34}~=~D_{q}(f_{2})_{43}=
D_{q}(e_{2})_{45}~=~D_{q}(f_{2})_{54}=[2]^{1/2}
\end{array} \eqno (4.2) $$

\noindent
The rest of elements are vanishing. Through tedious calculation, we
obtain:
$$\begin{array}{ll}
\left(L^{-}\right)^{1}_{~1}~=~t_{1}^{-1}t_{2}^{-2},~~~~~
&\left(L^{-}\right)^{2}_{~2}~=~t_{1}^{-1}t_{2}^{-1},\\
\left(L^{-}\right)^{3}_{~3}~=~t_{2}^{-1},~~~~~
&\left(L^{-}\right)^{4}_{~4}~=~1,\\
\left(L^{-}\right)^{5}_{~5}~=~t_{2},~~~~~
&\left(L^{-}\right)^{6}_{~6}~=~t_{1}t_{2},\\
\left(L^{-}\right)^{7}_{~7}~=~t_{1}t_{2}^{2},~~~~~
&  \end{array} \eqno (4.3) $$
$$\begin{array}{ll}
\left(L^{-}\right)^{2}_{~1}~=~-\omega_{2} e_{2} t_{1}^{-1}t_{2}^{-2},~~~~~
&\left(L^{-}\right)^{3}_{~2}~=~-\omega_{1} e_{1} t_{1}^{-1}t_{2}^{-1},\\
\left(L^{-}\right)^{4}_{~3}~=~-\omega_{2} [2]^{1/2} e_{2} t_{2}^{-1},~~~~~
&\left(L^{-}\right)^{5}_{~4}~=~-\omega_{2} [2]^{1/2} e_{2} ,\\
\left(L^{-}\right)^{6}_{~5}~=~-\omega_{1} e_{1} t_{2},~~~~~
&\left(L^{-}\right)^{7}_{~6}~=~-\omega_{2} e_{2} t_{1}t_{2},
  \end{array} \eqno (4.4) $$
$$\begin{array}{l}
\left(L^{-}\right)^{3}_{~1}~=~\omega_{2}
\left(q e_{1} e_{2}-e_{2} e_{1}\right)t_{1}^{-1}t_{2}^{-2},\\
\left(L^{-}\right)^{4}_{~2}~=~- \omega_{2} [2]^{1/2}
\left(e_{1} e_{2}-q e_{2} e_{1}\right) t_{1}^{-1}t_{2}^{-1},\\
\left(L^{-}\right)^{5}_{~3}~=~ \omega_{2}^{2}
q_{2}^{-1} e_{2}^{2} t_{2}^{-1},\\
\left(L^{-}\right)^{6}_{~4}~=~\omega_{2} [2]^{1/2}
\left(q e_{1} e_{2}-e_{2} e_{1}\right) ,\\
\left(L^{-}\right)^{7}_{~5}~=~- \omega_{2}
\left(e_{1} e_{2}-q e_{2} e_{1}\right) t_{2},
  \end{array} \eqno (4.5) $$
$$\begin{array}{l}
\left(L^{-}\right)^{4}_{~1}~=~\omega_{2} q_{2}^{2} [2]^{-1/2}
\left\{e_{1}e_{2}^{2}-([6]/[3])e_{2}e_{1}e_{2}+e_{2}^{2}e_{1}\right\}
t_{1}^{-1}t_{2}^{-2},\\
\left(L^{-}\right)^{5}_{~2}~=~- \omega_{2} [2]^{-1}
\left\{e_{1}e_{2}^{2}-q_{2}^{2}[2]e_{2}e_{1}e_{2}
+q_{2}^{4}e_{2}^{2}e_{1}\right\}
t_{1}^{-1}t_{2}^{-1},\\
\left(L^{-}\right)^{6}_{~3}~=~- \omega_{2} [2]^{-1}
\left\{q_{2}^{4}e_{1}e_{2}^{2}-q_{2}^{2}[2]e_{2}e_{1}e_{2}
+e_{2}^{2}e_{1}\right\}
t_{2}^{-1},\\
\left(L^{-}\right)^{7}_{~4}~=~ \omega_{2} q_{2}^{2} [2]^{-1/2}
\left\{e_{1}e_{2}^{2}-([6]/[3])e_{2}e_{1}e_{2}+e_{2}^{2}e_{1}\right\},
  \end{array} \eqno (4.6) $$
$$\begin{array}{l}
\left(L^{-}\right)^{5}_{~1}~=~\omega_{2} q_{2} [2]^{-1}
\left\{e_{1}e_{2}^{3}
-([4]-q_{2}^{-1})e_{2}e_{1}e_{2}^{2}+q_{2}([4]-q_{2})
e_{2}^{2}e_{1}e_{2}-q_{2}e_{2}^{3}e_{1}\right\}
t_{1}^{-1}t_{2}^{-2},\\
\left(L^{-}\right)^{6}_{~2}~=~- \omega_{1}\omega_{2}q_{2}^{2}[6]^{-1}
\left\{e_{1}^{2}e_{2}^{2}+[3]e_{2}e_{1}^{2}e_{2}-([6][5]/[3][2])
e_{1}e_{2}^{2}e_{1}+e_{2}^{2}e_{1}^{2}\right)t_{1}^{-1}t_{2}^{-1},\\
\left(L^{-}\right)^{7}_{~3}~=~-\omega_{2} q_{2}[2]^{-1}
\left(q_{2}e_{1}e_{2}^{3}
-q_{2}([4]-q_{2})e_{2}e_{1}e_{2}^{2}+([4]-q_{2}^{-1})
e_{2}^{2}e_{1}e_{2}-e_{2}^{3}e_{1}\right\}t_{2}^{-1},
\end{array} \eqno (4.7) $$
$$\begin{array}{l}
\left(L^{-}\right)^{6}_{~1}~=~\omega_{1} q ([6][2])^{-1}
\left\{ q_{2}^{2}e_{1}^{2}e_{2}^{3}+q_{2}^{-2}e_{2}^{3}e_{1}^{2}
+q_{2}([4]-q_{2}^{-1})e_{2}e_{1}^{2}e_{2}^{2}\right. \\
{}~~~~~+q_{2}^{-1}([4]-q_{2})e_{2}^{2}e_{1}^{2}e_{2}
-q_{2}^{2}([4]-q_{2})([6]/[3])e_{1}e_{2}^{2}e_{1}e_{2}\\
{}~~~~~\left.-q_{2}^{-2}([4]-q_{2}^{-1})([6]/[3])e_{2}e_{1}e_{2}^{2}e_{1}
+([6][4]/[3][2])e_{1}e_{2}^{3}e_{1}
\right\}t_{1}^{-1}t_{2}^{-2},\\
\left(L^{-}\right)^{7}_{~2}~=~\omega_{1} q ([6][2])^{-1}
\left\{ q_{2}^{-2}e_{1}^{2}e_{2}^{3}+q_{2}^{2}e_{2}^{3}e_{1}^{2}
+q_{2}^{-1}([4]-q_{2})e_{2}e_{1}^{2}e_{2}^{2}\right. \\
{}~~~~~+q_{2}([4]-q_{2}^{-1})e_{2}^{2}e_{1}^{2}e_{2}
-q_{2}^{-2}([4]-q_{2}^{-1})([6]/[3])e_{1}e_{2}^{2}e_{1}e_{2}\\
{}~~~~~\left.-q_{2}^{2}([4]-q_{2})([6]/[3])e_{2}e_{1}e_{2}^{2}e_{1}
+([6][4]/[3][2])e_{1}e_{2}^{3}e_{1}
\right\}t_{1}^{-1}t_{2}^{-1},\\
\left(L^{-}\right)^{7}_{~1}~=~\omega_{1}\omega_{2}
q ([6][2])^{-1}\left\{([6][4]^{2}/[3][2])e_{2}e_{1}e_{2}^{2}e_{1}e_{2}
-([6]/[3][2])e_{1}e_{2}^{4}e_{1}\right.\\
{}~~~~~\left.-([4][2]-1)e_{2}^{2}e_{1}^{2}e_{2}^{2}
+(e_{2}^{4}e_{1}^{2}+e_{1}^{2}e_{2}^{4})
-([6][5]/[3][2])(e_{1}e_{2}^{2}e_{1}e_{2}^{2}
+e_{2}^{2}e_{1}e_{2}^{2}e_{1})
\right\}t_{1}^{-1}t_{2}^{-2},
 \end{array} \eqno (4.8) $$

$\left(L^{+}\right)^{a}_{~b}$ can be obtained from
$\left(L^{-}\right)^{b}_{~a}$  by following transformations: replace
$q_{j}$ by $q_{j}^{-1}$ (so $\omega_{j}$ becomes $-\omega_{j}$),
$e_{j}$ by $f_{j}$, $t_{i}$ by $t_{i}^{-1}$, and reverse the
product order of operators in each term. For example,
$$\left(L^{+}\right)^{1}_{~4}~=~-\omega_{2} q_{2}^{-2} [2]^{-1/2}
t_{1}t_{2}^{2}
\left\{f_{2}^{2}f_{1}-([6]/[3])f_{2}f_{1}f_{2}+f_{1}f_{2}^{2}\right\}$$

\noindent
It seems to us that this relation between $\left(L^{+}\right)^{a}_{~b}$
and  $\left(L^{-}\right)^{b}_{~a}$ holds for all quantum
enveloping algebras.

\vspace{10mm}
{\bf References}

\noindent
[1] V. G. Drinfel'd, Quantum Group, Proceedings of the International
Congress of Mathematicians, Berkeley, 1986, Vol.1, p.798.

\noindent
[2] M. Jimbo, Lett. Math. Phys. {\bf 10}(1985)63; Commun. Math. Phys.
{\bf 102}(1986)537.

\noindent
[3] M. Jimbo, Topics from representations of $U_{q}(g)$, - An
introductory guide to physicists, 1991, Nankai Lectures on
Mathematical Physics, World Scientific, Singapore, p.1.

\noindent
[4] Zhong-Qi Ma, Yang-Baxter Equation and Quantum Enveloping Algebras,
World Scientific, Singapore, 1993.

\noindent
[5] C. N. Yang, Phys. Rev. Lett. {\bf 19}(1967)1312; R. J.
Baxter, Ann. Phys. {\bf 70}(1972)193.

\noindent
[6] E. Date, M. Jimbo, K. Miki, and T. Miwa, Commun. Math.
Phys. {\bf 138}(1991)393.

\noindent
[7] M. Jimbo and T. Miwa, Algebraic Analysis of Solvable
Lattice Models, Conference Board of the Mathematical
Sciences, Regional Conference Series in Mathematics,
No. 85, 1995.

\noindent
[8] T. Brzezinski and S. Majid, Commun. Math. Phys. {\bf 157}(1993)591.

\noindent
[9] S. Watamura, Commun. Math. Phys. {\bf 158}(1993)67.

\noindent
[10] Bo-Yu Hou, Bo-Yuan Hou and Zhong-Qi Ma, J. Phys. A. {\bf 28}(1995)543.

\noindent
[11] L. D. Faddeev, N. Y. Reshetikhin and L. A. Takhtajan,
Quantization of Lie groups and Lie algebras, in Algebraic
Analysis, Academic Press, 1988, p.129.

\noindent
[12] Ke Wu, Han-Ying Guo and Ren-Jie Zhang, High Ener. Phys. and
Nucl. Phys. {\bf 17}(1993)262 (in chinese); Ke Wu and Han-Ying Guo, High
Ener. Phys. and Nucl. Phys. {\bf 17}(1993)796 (in chinese).

\noindent
[13] P. Schupp, P. Watts and B. Zumino, Commun. Math. Phys.
{\bf 157}(1993)305.

\end{document}

redpajama_set_name: RedPajamaArXiv


Well, it is a Hall, n'est pas, and it's er, famous (locally).
It therefore qualifies on at least two counts.
Off Burnley Road is Huncoat Hall of late medieval origin, this was the home of the Birtwistle family of Huncoat, the first of which was William of Birtwistle who was born in 1275 and died in 1330.
The name Birtwistle runs like a thick thread through the historical tapestry of East Lancashire's now virtually-vanished cotton industry.
Reaching its peak in the early 1930s, the empire built by Blackburn-born William Birtwistle extended to 17 mills in his home town, Preston, Darwen, Great Harwood and Abbey Village, and in addition to that vast number of looms it also included 350,000 spindles.
Though it lasted 150 years from when its foundations were laid by his grandfather, nothing now remains of that giant industrial realm -- one that employed 9,000 people and enabled the head of the William Birtwistle Allied Mills Group to acquire huge mansions and own giant steam yachts.
Huncoat Hall was in the possession of Mr. Foot in 1825, but later acquired by the Towneleys, of Towneley Hall, Burnley, and in 1839 was in the occupation of Henry Sudall. Later the Hall was occupied by the Haworth family for many years. The last of the male line of the Towneleys left two daughters who married Lord O'Hagan and Lord Abingdon. When the estates were divided Huncoat Hall estate passed into the possession of Lord Abingdon, who sold it to Mr. James Crook, the present owner and occupier, a member of the last Huncoat Parish Council previous to amalgamation with Accrington in April, 1929.
Anbody know anything about Birtwistle's Ice Creams ?
I thought it was Birtwell's Ice Cream van that use to come round!
No, wasn't that Tricketts from Feilding Lane?
The Proctor family worked Huncoat Hall in WWII, and after.
Is there any threads on them?
Any family information on the Jacksons from High Riley Farm?
While on farms and farmers, 'Hillock Farm', at the bottom of the Coppice.
We used to sledge in a field below, closer to Accy Stanley.
Talking of Huncoat, would anyone know where I might find a copy of HUNCOAT REMEMBERED The recollections of John Smith.
I think it was published in 1985 by the Hyndburn Local History Society.
I can't seem to find a copy anywhere.
My landlord is called Jackson, and still lives at High Riley.
Hey !! my names Laura Jackson and i live at High Riley !!! What do you want to know about us ???
checked a bit for you Beechy, looks like Sarah Ann Pilkington married Lawrence Dewhurst in 1880 at St James Church Kirk. They had moved to Huncoat Cottage by 1881 which is/was next to Huncoat Hall, Lawrence as a farm labourer. Lawrence was born in Padiham and Sarah born Dukinfield Cheshire, they still lived in the cottage in 1901 but had no children that I could find. Not much of a tale but that,s all I could come up with.
I used to stay at the farm when my Aunty Glady's [ ne Proctor] lived there.
I am a direct descendent of Oliver Birtwistle who resided at Huncoat Hall.
Who lives there currently and does anyone have any photos.
On the other point yes Harrisons music is unique and an aquired taste.
It has made him a lot of money.


text:
Tällberg Foundation
Doing Nobody's Job
by Sara Ersson | Oct 4, 2019 | Leadership | 0 comments
Curiosity. Audacity. Determination. A bit of heart. That how Jaykumar Menon describes the qualities that have propelled him, as he followed his teenage ambition "to positively affect the lives of tens of millions of people." A lofty goal, which is now within reach. Five years ago, this international human rights lawyer, scholar, and social entrepreneur—and finalist for this year's Tallberg/Eliasson Global Leadership award—launched what is now the Open Source Pharma Foundation, to make new instantly generic drugs.
"The medicines that we have are really driven by a very particular market system," Menon notes. "Only five percent of human diseases have an FDA-approved cure right now. There's just not the market incentive to address the majority [of them]…. And because of that billions of people are really left in the lurch." Using open source principles, crowdsourcing, computation—in some cases AI—and a grassroots network of intellectual production, he is constructing an alternative approach.
With pride, he reports a big early victory: "We are in late stage clinical trials for a new therapy for tuberculosis and rather than spending $100 million, $500 million, which one might in a Big Pharma setting, we got there to the beginning of phase 2B literally off of my credit card—and it's not a very large credit card," he adds, with a smile. "So it was just a clever use of open IP and open principles."
Menon began his legal career by exonerating a man falsely convicted of murder. He was the fifteenth lawyer to try to free David Wong. His solution to what seemed an unwinnable case? Find the real killer. It took him five years, and meant he had to quit his paying job, but he would not contemplate failure. In the public health arena, he is pushing another radically simple solution, to the widespread nutrition problem of iron deficiency. "Iodized salt reaches five billion people," he says. "Two billion people don't have enough iron, and we're adding iron to salt." The doubly fortified salt is already reaching over 10 million people.
For all his credentials—J.D. and master of international affairs from Columbia, a year of medical school at Brown, professorship at McGill's Institute for International Development, funding from the Gates and Rockefeller foundations, etc.—Menon often operates as an non-expert, working "from first principles," as he puts it. "These are outsider approaches, but it's not rocket science. You can kind of understand some of the basics. You ally yourself with the experts and what's missing is—it's nobody's job to solve world hunger. It's nobody's job to create affordable medicines. And sometimes just what's needed is a vision and some skillful sheep-herding and pushing."
Menon has many visions—"a sort of sea of ideas," he calls it—such as the "SDG Olympics," which would reward and glorify countries for reaching sustainable development goals. What keeps him from getting sidetracked is his methodology: "It's to look at a problem that's A) serious, B) affects a billion or more marginalized or underprivileged people, C) is subject to some kind of transformative change—it's important to pick your spots wisely—D) through some kind of open innovation technique." Those criteria satisfied, Menon's job of sheep-herding and pushing begins.
Names of the Dead
Into Africa: the Tällberg Foundation Celebrates Great Leadership in Kenya
New Thinking for a New World
Tallberg / Eliasson Global Leader
info@tallbergfoundation.org
Tällberg / Eliasson Global Leadership Prize ©2018 | Powered by Narrative Studio


text:
Luol Deng could return to Bulls lineup Sunday against Heat
By Kurt HelinJan 26, 2012, 8:35 AM EST
When the Bulls said earlier this week that Luol Deng would be out "for a while" what they meant was less than a week. Apparently.
Deng has torn ligaments in his wrist but could return to the lineup Sunday when the Bulls have a big showdown with the Miami Heat, reports Adrian Wojnarowski of Yahoo.
The Chicago Tribune reported that Deng has been telling teammates he plans to play Sunday, but the organization has been more cautious about his outlook. Doctors told Deng he ultimately needs surgery on his left wrist – which isn't his shooting hand – and that it could sideline him for three to four months.
It's like an organization to be cautious, especially this early in the season, but the Bulls miss Deng's versatility (they could have used him Wednesday against the Pacers). He is the second leading scorer on the team but he also brings rebounds, defense and more — he is a glue guy. Somebody who helps keep all the parts of the Bulls together.
They're going to need guys like that against the Heat.
Not sure when he is going to have that surgery — right after the season Deng is expected to play for host Great Britain in the Olympics. Surgery after that means he misses training camp and likely the beginning of next season.


text:
/**
 * This file exists to provide an example of how to use the Jest module. It has nothing to do with the functionality of the overall project.
 */
import { sum } from './sum';

test('adds 1 + 2 to equal 3', () => {
  expect(sum(1, 2)).toBe(3);
});

redpajama_set_name: RedPajamaGithub


We're excited for Summer, so excited we've come up with our best Summer menu yet! There's something for everyone, from sweet to savory to spicy. The menu features 2 new housemade sauces AND slow braised carnitas.
We have two new specials for a limited time only- March 1st-April 13th.
Summer has finally arrived and so have our seasonal specials. Come in and try our delicious Summer menu, from a refreshing cup of iced coffee to a savory Chipotle Chicken Quesadilla. Available for a limited time only!
Last Chance To Enjoy Our Comfort Classics!
It's time to say goodbye to Old Man Winter and say hello to Spring. Our delicious Comfort Classics are bidding adieu March 13th, so come in to get your last yummy bite of the popular Low Country Benedict!


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My philosophy is tinted moisturizer/BB cream for daytime, and a medium to full coverage foundation sans SPF for nighttime and special occasions. Lately, I have been completely obsessed with trying every tinted moisturizer and BB cream I could get in my makeup addicted hands. I love how easily I can slather tinted moisturizers and BBs with my hands for a quick FOTD. The ease of going from light to medium coverage is also a perk.
The Smashbox Camera Ready BB came as a sample in a recent Sephora haul. Smashbox's BB provides SPF 35, and is available in 8 different shades. Generally, for a no makeup makeup look, I only need about a pea-size for my entire face. I think the 1 oz. size would last me quite a while, probably 3 or 4 months. Sephora offers two shades in .5 oz travel sizes -- "Light" and "Light/Medium". Although, I was sent 'Light" instead of "Light/Medium", I was still able to make it work after bronzing and blush. *For reference, I am NC25 in MAC.
I admit that I totally underestimated how much I would like the Smashbox BB. I was pleasantly surprised at how long it lasted -- a good 6-7 hours. Although the matte finish formula is thicker than my watery Nars Pure Radiant Tinted Moisturizer (which I wear in St. Moritz), it worked surprisingly well with my dry skin. To the naked eye, the matte finish does make you look "Camera Ready", but I would look for something with more coverage for a super important day, like a wedding or school pictures.
On the daily, I prefer a more demi-matte finish so, I have been mixing Smashbox's BB with my super dewy Nars Tinted Moisturizer. I have actually noticed longer staying power with my Nars by mixing the two. I am thinking about ordering a "Light/Medium" and a lighter shade of my Nars tinted moisturizer to mix for the upcoming Fall/Winter. This is definitely a must try. Available at Sephora, Nordstrom, ULTA, Macy's, and other retailers for $39.00 1oz./$19.50 .5oz.
Have you tried the Smashbox Camera Ready BB?


text:
package com.github.pfichtner.jrunalyser.base.data;

import java.io.Serializable;
import java.util.concurrent.TimeUnit;

import com.google.common.primitives.Doubles;

public class DefaultPace implements Pace, Serializable {

	private static final long serialVersionUID = -82131255646159944L;

	private final double tPerD;
	private final TimeUnit timeUnit;
	private final DistanceUnit distanceUnit;

	private static final long fix = TimeUnit.MILLISECONDS.convert(1,
			TimeUnit.DAYS);

	public DefaultPace(double tPerD, TimeUnit timeUnit,
			DistanceUnit distanceUnit) {
		this.tPerD = tPerD;
		this.timeUnit = timeUnit;
		this.distanceUnit = distanceUnit;
	}

	protected static DefaultPace of(Speed speed, TimeUnit timeUnit,
			DistanceUnit distanceUnit) {
		return new DefaultPace(1 / speed.getValue(distanceUnit, timeUnit),
				timeUnit, distanceUnit);
	}

	@Override
	public DistanceUnit getDistanceUnit() {
		return this.distanceUnit;
	}

	@Override
	public TimeUnit getTimeUnit() {
		return this.timeUnit;
	}

	@Override
	public double getValue(TimeUnit timeUnit, DistanceUnit distanceUnit) {
		return convert(timeUnit, distanceUnit).tPerD;
	}

	public DefaultPace convert(TimeUnit timeUnit, DistanceUnit distanceUnit) {
		if (distanceUnit == this.distanceUnit && timeUnit == this.timeUnit) {
			return this;
		}
		long longDist = (long) (fix * this.distanceUnit.convert(this.tPerD,
				distanceUnit));
		return new DefaultPace((double) (timeUnit.convert(longDist,
				this.timeUnit)) / fix, timeUnit, distanceUnit);
	}

	@Override
	public Speed toSpeed(DistanceUnit distanceUnit, TimeUnit timeUnit) {
		return new DefaultSpeed(1 / getValue(timeUnit, distanceUnit),
				distanceUnit, timeUnit);
	}

	@Override
	public String toString() {
		return this.tPerD + " " + this.timeUnit + " per " + this.distanceUnit;
	}

	public int compareTo(Pace o) {
		return 0 - Doubles.compare(getValue(this.timeUnit, this.distanceUnit),
				o.getValue(this.timeUnit, this.distanceUnit));
	}

}


LUXEMBOURG: November 29, 2017. Cargolux has signed a new code-share and capacity swap agreement with Nippon Cargo Airlines (NCA) between Europe and Japan with effect from April 2018.
Cargolux will now be able to utilize NCA's flights from Frankfurt Hahn to Narita while NCA has reciprocal access to Cargolux's flights from Luxembourg.
As a result, Cargolux plans to introduce a weekly flight from Luxembourg to Tokyo Narita and Cargolux Italia will stop flying to Narita but continue its services to Osaka's Kansai airport.
In October 2015, Cargolux Italia and NCA signed a cargo codeshare agreement for flights from Milan Malpensa to Kansai and Narita. This was in addition to two weekly flights to Osaka, four from Milan Linate to Narita plus three weekly Cargolux flights from Luxembourg to Komatsu - which the airline says it now plans to develop further.
With cheaper landing fees and a 3,000 meter runway, Komatsu is only six hours by truck from Tokyo, four hours from Osaka and around three hours from Nagoya. The airport operates a 4,000 square meter bonded warehouse together with Nippon Express, Kintetsu and Yusen.


text:
Travel|Oil Money Flows, and City Prospers
Travel | Emerging Destination | Almaty, Kazakhstan
Oil Money Flows, and City Prospers
By ILAN GREENBERG DEC. 9, 2007
A soldier on steps leading to the Medeo winter sports complex in the Tien Shan mountains south of Almaty, a popular weekend spot. Credit Dean C.K. Cox
ON most Friday nights in Almaty, the Uzbek-themed night spot known as the Car Wash — an ornately decorated rooftop restaurant with enviable mountain views — is packed with well-heeled Kazakhs smoking water pipes, drinking, dancing and eating extravagantly. Sandwiched between a residential district scheduled for rapid development and Almaty's business district, the Car Wash is not the only center of hyperactive nocturnal activity.
Infused with newly flowing oil money, Kazakhstan's largest city is flush with nightclubs and exotic restaurants. The city's main boulevards are lined with English-language signs; boutiques sell everything from Armani to gem-encrusted Vertu cellphones, and cafes serve the latest in overpriced coffee concoctions.
Sacha Baron Cohen's comic Borat character may have imprinted in the minds of many people a sense of ridiculousness about Kazakhstan, but there is little that is ridiculous about this sprawling business hub. Half a dozen luxury hotels are planned or under construction, including an ambitious JW Marriott Hotel opening next summer. The country's flagship carrier, Air Astana, has added international flights to cities like Hanover, Germany; Dubai; and Bangkok. Wide-bodied Airbus and Boeing jets have joined its fleet.
In short, Almaty is no longer a hardship outpost for the diplomats and the oil industry executives who still dominate the city's visitor logs. Although the overwhelming percentage of visitors continues to be businesspeople, some have begun extending their stays to spend a weekend skiing above the Medeo winter sports complex, to visit nearby nature preserves (among them, the Charyn Canyon, which in breadth and sheer drama compares favorably to the Grand Canyon), or just to explore the city of nearly two million people. "Almaty feels raw, sketchy, and that can be fascinating," said Brooke Arnao, the online video director for Money-Media, a New York publishing company, who stayed for a weekend after a conference.
But although the forested streets and urban parks create pleasant, verdant scenes in summer, the city has little of the Silk Road charm of neighboring cities like Samarkand and Bukhara in Uzbekistan. Much of Almaty's architecture dates from the 1960s and '70s, not a celebrated time in Soviet design. Still, there are several worthwhile city pilgrimages. The Russian Orthodox church in Panfilov Park, built entirely of wood — including the nails — sits across from the Artisan bathhouse, an opulent Soviet vision of the old take-to-the-waters remedy. The Almaty art museum has fine examples of contemporary Kazakh art, and the carpet sellers on the ground floor sell Kazakh rugs (a 4-by-10-foot carpet runs about 73,000 tenge, or $594, at 125 tenge to the dollar.)
But Almaty's emergence as a tourist destination is just beginning. There are few "official" taxis. Some museums are dilapidated. Many local tour operators, like David Berghof of Stantours, focus on the city as the best gateway to a Silk Road tour. "Almaty has become a convenient starting point," he said, "with a well-functioning infrastructure."
For those who do stick around, encounters with the city's polyglot population will bring them in contact with the legacy of the Soviet gulag — many of Almaty's residents are the descendants of the sprawling prison system, a good part of which was built in Kazakhstan. Striving immigrants from other Central Asian countries work in cafes and stores, and the city's restaurants reflect its diverse population. Fragrant Uighur cuisine is informed in equal parts by Uighur's roots in western China and its ethnic connections to Turkey. Numerous restaurants offer excellent interpretations of ethnic Russian and Ukrainian cuisines, and a lamb and noodle dish called beshbarmak is a good introduction to the country's more regional nomadic kitchen.
Meanwhile, the bulk of Kazakhstan's oil reserves have yet to be tapped. What will this incipiently wealthy country look like in 10 years' time — the Dubai of Central Asia, perhaps?
A version of this article appears in print on , on Page 56 of the New York edition with the headline: Oil Money Flows, and City Prospers. Order Reprints| Today's Paper|Subscribe


Any and all jewelry are always the perfect gift when in doubt. If you have not seen the Gypsy jewelry box yet, visit us! We carry everything from nice silver, to midi rings, to big turquoise, to dainty gold necklaces.
Living in the boutiful Willamette Valley, we Oregonians love our fresh produce. And when the farmer's markets are not open in the winter, where else do we get our fresh basil? The kitchen counter, of course! Soil-free Aerogarden by MiracleGrow: each pod contains leafy produce of your choice, which begins to sprout in 4-8 days. You can harvest the same plant for up to five months. Now this is the gift that keeps on giving, Clark.
Do we need to explain the Man hotdog roaster?
Got hot coffee? You will still have coffee at dinner. Iced tea? It will stay iced until tomorrow.
Remind her that she is special to you with dainty jewelry that she can look at and show off every day.
This magical wrap can be worn as a skirt, a scarf, a vest and more! It's perfect for those crazy Oregon weather days when it's sunny in the morning and chilly in the afternoon.
Whether you're on the go a lot or just forget to charge your phone a lot, the ChargeHub is perfect for charging multiple USB plug devices at once. Can charge seven devices at the same time.


text:
Q: How add Third UIView in UISplitViewController? Can and how I add UIView above masterView in UISplitViewController.
see example:

A: If you want something more advanced than what Novarg has suggested, use MGSplitViewController. You can then have a split view controller as the detail view of the top split view controller. MGSplitViewController also allows for resizable ratio so you can fit it as you like.


text:
.montage-Flow {
    position: absolute;
    top: 0;
    left: 0;
    bottom: 0;
    right: 0;
    overflow: hidden;
    -webkit-transform: translateZ(0);
    -moz-transform: translateZ(0);
    -ms-transform: translateZ(0);
    transform: translateZ(0);
}

.montage-Flow-camera,
.montage-Flow-Repetition {
    position: absolute;
    top: 0;
    left: 0;
    bottom: 0;
    right: 0;
    -webkit-transform-style: preserve-3d;
    -moz-transform-style: preserve-3d;
    -ms-transform-style: preserve-3d;
    transform-style: preserve-3d;
    -webkit-transform: translateZ(0);
    -moz-transform: translateZ(0);
    -ms-transform: translateZ(0);
    transform: translateZ(0);
}

.montage-Flow-Repetition .no-transition,
.montage-Flow-Repetition .no-transition * {
    -webkit-transition: none !important;
    -moz-transition: none !important;
    -ms-transition: none !important;
    transition: none !important;
}

.montage-Flow-Repetition > div {
    position: absolute;
    top: 50%;
    left: 50%;
    width: 0;
    height: 0;
    -webkit-transform: translateZ(0);
    -moz-transform: translateZ(0);
    -ms-transform: translateZ(0);
    transform: translateZ(0);
    -webkit-transform-style: preserve-3d;
    -moz-transform-style: preserve-3d;
    /* Does not work: http://msdn.microsoft.com/en-gb/library/ie/hh673529%28v=vs.85%29.aspx#the_ms_transform_style_property */
    -ms-transform-style: preserve-3d;
    transform-style: preserve-3d;
}

.montage-Flow-Slot {
    position: absolute;
    top: 50%;
    left: 50%;
    -webkit-transform: translate3d(-50%, -50%, 0);
    -moz-transform: translate3d(-50%, -50%, 0);
    -ms-transform: translate3d(-50%, -50%, 0);
    transform: translate3d(-50%, -50%, 0);
}


Finally the BOOT has gotten off the lazy boy relaxation approach to marketing with a ground breaking URL marketing strategy. After battling with GoDaddy for about an hour I managed to secure http://www.thebusinessofonlinetravel.com/. I even manage to figure out how to do a redirect to the blogger address. One day I may even generate the courage to migrate the blog to Wordpress/SquareSpace, host the thing on some dedicated servers and get really serious.
yourURL/%labelname/%postname(instead of yourURL.com/%year/%month/%postname) for SEO purposes.
Pretty easy to do in WP not sure about Blogger or SqSpace.
You just need to make sure you never have two posts with the same headline. Or you will break the Intertoobs and the world will end.
In any case, as I imagine Google indexes your blog pretty well, it will just be a improvement in the margin - but that is the case with most SEO.
@Carl - thanks for the tip. I know that I have to get off blogger and onto a more flexible platform. Not only for SEO purposes but so I can add pieces like "top posts of the week", "commentary feeds", sharing on digg/delicious etc. Problem is I am a techno-wannabe who is scare of screwing up a migration and losing three years of page rank.


The façade of 801 Grand has been reenergized. The iconic skyscraper was outfitted with new high-efficiency, low-maintenance LED fixtures that emit a whiter, brighter light, illuminating perhaps the most recognizable building in the Des Moines skyline from top to bottom. It is the first time in more than a decade the entirety of the building has been lit.
This update comes as part of a $284 million renovation of the Principal Financial corporate headquarters campus in downtown Des Moines, a 7-year project for which OPN Architects served as the master planner and primary architect. The project includes the renovations of three buildings and three skywalk bridges, and exterior improvements to several buildings on the campus. Check out more coverage and images of the project on the Des Moines Register.


Best Punjabi Status for Whatsapp to Impress friends, lover, GF & BF with funny, lovely, stylish, cool & cute quotes in Punjabi.
Toot jaanda hai gareebi vich oh rishta jo sabto khaas hunda hai, Ban jaande hazaara yaar usde paisa jad paas hunda hai.
One might know many language and can use English for impression but your mother tongue and pind is the only thing that you love most, each and every expression in foreign language sounds fake also emotions less.
Our love for our Punjabi language is priceless, so for our friends from Punjab we are sharing Punjabi status for whatsapp, ow you can put these Punjabi status on your whatsapp DP to impress your friends, lovers, girlfriend, and boyfriend. The most interesting thing of our latest Punjabi whatsapp status collection is its versatility, you can find funny, cool, love, tashan, cute, emotional and sad Punjabi status under one roof.


text:
Q: Swift Array/Dictionary is not printing back in expected order Im attempting to to combine two swift arrays into a simple swift dictionary. I am extremely new to swift (but know a decent amount of python), and have come across something that I think is rather odd. When I run the code below, I get a dictionary just as expected, however, if I rerun the code, the key:value pairs come up in a different order. I know that dictionaries in swift are supposed to be unordered, but does this mean each time I go to iterate through the key:value pairs of a dictionary, the order of these pairings may change? In other words, is it impossible to preserve the order to the two initial arrays that comprise the dictionary in the final dictionary product?

let cities: Array<String> = ["Paris", "Mexico City", "Rome"]
let countries: Array<String> = ["France", "Mexico", "Italy"]

var newCountrydict : Dictionary<String, String> = [:]
for (city,country) in zip(cities,countries){
    newCountrydict[country] = city

}

newCountrydict

A: In the swift docs, under "Dictionaries":

Unlike items in an array, items in a dictionary do not have a specified order. You use a dictionary when you need to look up values based on their identifier, in much the same way that a real-world dictionary is used to look up the definition for a particular word.

It is not possible to preserve the order if you use dictionaries in swift.
In the same docs, it says:

Arrays are ordered collections of values. Sets are unordered collections of unique values. Dictionaries are unordered collections of key-value associations.

I suggest you read through the official documents.


Edmund Gregory Hinkson is a Barbadian politician and a current Cabinet Minister.
Edmund Gregory Hinkson was educated at then New Pine Primary School, Harrison College, Cave Hill Campus of the University of the West Indies earning a Bachelor's of Laws degree with Honours, a Certificate in Legal Education and a Master's in Business Law. Added to these were many years of community service through the Lions Club, St. John Ambulance Association, National Crime Prevention Panel, Advisory Board of the Salvation Army, and service to the disabled community.
The attorney-at-law gained governmental experience as Chairman of the National Conservation Commission and the Severance Payments Tribunal; Deputy Chairman of the National Housing Corporation. National Advisory Committee on Disabilities and the Building Advisory Committee; a Director of the National Cultural Foundation; and Member of Consumer Claims Tribunal, Income Tax Appeal Board and the Harrison College Board of Management.
Edmund was elected to the Barbados House of Assembly in the 21 February 2013 General Election.
Hinkson retained his seat in the 24 May 2018 election and was appointed Minister of Home Affairs in the Mia Mottley Adminsitration.
Hinkson is married to businesswoman Beverly and they are the parents of Erica and Gregory. His non-work interests are playing lawn and table tennis and listening to jazz.


text:
FRIDAY DINNER WENT WELL
Thank you for praying for our monthly dinner held at Lutheran Church of The Cross last week. There were 60 in attendance. Jan Holter (left) led us in anointed praise and worship. Carl Holter serves on our Board of Directors and Jan is an accomplished pianist. The Holters live in Wisconsin. Carrie Irving (left), Adviser to the Board, shared a dynamic testimony of the Valley Vineyard's ministry in the mountains of Oaxaca. Everyone was enthralled.
PEREAU'S SPEAKING ENGAGEMENTS
This Saturday and Sunday, the Pereaus will be ministering at Gloria de Cristo Church, 11273 E. 40th Street, Yuma, AZ 85367. Service times are—Saturday 3 pm and 5 pm; Sunday 7:45 am, 9:30 am, and 11:00 am. Duane Weeks will be hosting the Pereaus. They will also revisit the Gretna Green wedding chapel where they were married 68 years ago.
PASTORS CONFERENCE IN OAXACA
A team from our Baja and Oaxaca Missions hosted hosted a pastors conference in the mountains of Juxtlahuaca last weekend. Mario Cordoba was one of those who was there to minister. Praise God, pastors and missionaries were blessed beyond measure.
HOUSE MOTHERS NEEDED
We still have a need in both Baja and in Oaxaca for house mothers. Pray for women who have the capacity to love the children of others.
TEACHER NEEDED IN OAXACA
A bilingual elementary school teacher is needed in Oaxaca. And, of course, we continue to pray for a doctor for our medical center in Baja.
MOUNTAIN MISSIONARIES
Please pray for our mountain missionaries, Maria and Teresa.They have been given the challenge from Samaritains Purse to take gifts to an unreached community.This Saturday (February 11th) they will be going to the community of Encino Amarillo to have this activity. Please also pray for safe travels, as Maria's vehicle has been having mechanical problems recently.
The little red Nissan truck of Maria's has died. She needs a replacement at approximate cost of $16,000. We have received $9,100 towards the truck. You may give toward this project here… https://www.ffhm.org/support-a-project/truck-for-maria-in-oaxaca/
PAT DURKIN
Pat Durkin will return to her new residence this weekend. She has been given a clean bill of health from her doctor. Praise God.
SHERLYN
Sherlyn, a student at the TJ House, had to have emergency surgery Thursday morning to remove a large mass from her abdomen. She is still in the hospital recovering, and the mass was sent to pathology to make sure it wasn't cancerous. Please pray for her quick and cancer-free recovery.
JIM KNOWLTON
Jim Knowlton, former missionary to Papua New Guinea and school teacher, had a problem swallowing. After some time, paramedics were called and he was taken to the hospital. The doctors detected a more serious problem and said that if he had not been taken to the hospital, he would have died. What the enemy meant for evil God turned to good. His wife, Bertie Knowlton, now deceased, did the illustrations for The Little Lost Pup, available from the US office.


Anzac Day has become a fascinating paradox.
The numbers of Gallipoli veterans have almost entirely dried up. Hardly any First World War ex-servicemen are left. The ex-service community of the Second World War is also diminishing. One might have thought that Anzac Day was destined to disappear.
Yet somehow, younger generations born after all of Australia's wars have begun to discover the day and to observe it with interest, fascination and even pride. Anzac Day, it seems, is back! Anzac Day is alive again!
So much so that there are suggestions that this might, and should, become the national day and replace the twenty-sixth of January because the symbolism of Anzac Day is superior.
What is the argument in favour of Anzac Day?
It symbolises Australian courage and striving. It commemorates the discovery of national identity and character. It resounds with colour, charisma and legend. It includes the Aboriginal and not only the white community. For newer Australians it is seminal in the Australian ethos which they have made their own. It enables a national day to begin with solemnity and move into celebration.
From the practical point of view, it is a better time of year than the end of January, when many people are still on holiday. And 26 January itself has not always been such a major occasion; until relatively recently it was known by other names and lacked major significance.
What statement would Anzac Day make as Australia's national day?
Critics accuse the day of glorifying or mythologising war. Which is nonsense. No-one believes war has any glory; no-one suggests that is what Anzac Day is about. Ask anyone who was ever on active service.
No; Anzac Day symbolises Australian identity based on mateship, even a streak of larrikinism, and on doing something for your country. Australia needs both these concepts today.
It needs mateship. No-one can be proud of the inter-community, inter-ethnic tensions that have become a worrying sign of the Australian scene. Older Australians prejudiced against newer? Aboriginal Australians denied dignity by non-Aboriginals? Migrant groups who import the conflicts that have torn their homelands apart? There is not much Australian mateship there.
And what about the growing insecurity of every Australian whatever their background? Suburban streets that are unsafe to walk? Peaceful homes that are almost fortresses? Australians frightened of other Australians? Yes, other countries are far worse, but we do not need these manifestations here on any level. We need mateship and fair-go.
Anzac Day also stands for doing something for your country, exerting yourself to celebrate, enhance and enrich it. Instinctively. Lovingly. Proudly.
Rorting the system is out. The world does not owe you a living! Corruption is out. There have to be standards!
Doing something for your country means bringing your talents, energies and interests to bear in solving national problems, promoting national ideals, creating a climate of good citizenship.
When you go overseas as an Australian you come back feeling good about Australia. It is not only that here there is space and the sun shines. It is because of the ethos and tradition which Anzac symbolises, and because we have created a way of life that has no equal anywhere.
That's what we need to say on whatever date is our national day. We need to say it most of all to ourselves, that we have something precious here, and we have to cherish, appreciate, celebrate, preserve, protect and strengthen it. We have to ensure that no-one, for any reason, is allowed to undermine Australia and Australianism.
That is what the ex-service community have always known. That is what Anzac Day has always represented. As far as I am concerned, I hope it will become our national day. But whether that happens or not, it is good to see the day rediscovered.
This article first appeared in April, 1996.


text:
HomeCraighead CountyMentally challenged man leads police on chase with bad ending
Mentally challenged man leads police on chase with bad ending
Nov 11, 2016 NEA Report Craighead County, Poinsett County 0
A number of JPD patrol cars parked near the Justice Complex on Washington in Jonesboro.
JONESBORO, Ark. – A man who was reported missing was later discovered in a nearby county, chased by authorities and wrecked a vehicle in a crash which sent him to The Med in Memphis.
The initial report began after 6:30 a.m. Thursday morning, as Jonesboro Police Department were told of an 18-year-old who was mentally challenged and had left the residence the night before, around 10 p.m., never to return. A relative of the man said he was angry because she wanted him to go live with their mother in Mexico. She also said he has mental problems.
Shortly after 8 p.m. the man was located in Poinsett County, according to their sheriff's department. A deputy contacted JPD and informed officers that the Poinsett County Sheriff's Department had been in a chase with the man, who had stolen a vehicle at the Poinsett Package Store near the county line on Highway 1.
The deputies pursued the man in the stolen vehicle all the way to Wynne, where he wrecked the vehicle, the report says. He was airlifted to The Med in Memphis by Hospital Wing.
Old Country Store staying in business and staying quiet
Suspected bank robber arrested by TPD
Three New Exhibitions to Open This Week in Bradbury Art Museum
Sep 7, 2022 NEA Report Community, Craighead County, Email, Top Story 0
JONESBORO – Three new exhibitions will open in Bradbury Art Museum (BAM) at Arkansas State University with a 5-6:30 p.m. reception Thursday, Sept. 8. "People/Places/Things: Perspectives on Identity and Moments of Familiarity" brings together 14 artists, from nine […]
Greene County jury slams child molester with six life sentences
Dec 14, 2018 NEA Report Crime, Greene County, Top Story 0
Two defendants convicted on rape counts of a juvenile in Greene County. […]
City, partners seek $24MM federal connectivity grant
Jun 24, 2021 Stan Morris Community, Craighead County, Top Story 1
Jonesboro Mayor Harold Copenhaver announced Thursday that the City, in partnership with Arkansas State University and with the support of the region's Congressional delegation and other leadership, will apply for $20 million in funding that […]