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Home » World News » Furlough replacement: How does the Job Support Scheme work?
Furlough replacement: How does the Job Support Scheme work?
Rishi Sunak's furlough replacement: How does the Job Support Scheme work, who is eligible and how much will I get?
The Chancellor unveiled the Job Support Scheme (JSS) this afternoon
It will 'directly support' wages of staff in 'viable' roles for six months from Nov 1
Will involve firms paying a greater share of wages for hours not worked
Economists warn that it will not prevent a large number of job losses this winter
Rishi Sunak's new help for business and jobs at a glace
Furlough scheme will be replaced with a Job Support Scheme (JSS) to 'directly support' wages of staff working at least a third of regular hours.
Firms will pay workers for hours they work. For regular hours they cannot work, employer and Treasury will both pay a third of their wages, so they get two-thirds of pay for missed hours.
JSS open to firms who have not used furlough.
Self Employment Income Support Scheme Grant (SEISS) is extended, with a lump some to cover November to January next year
It will be worth 20 per cent of average monthly profits, capped at £1,875.
Second grant available for February to April 2021
Bounceback Loan guarantee term extended to 10 years from the current six.
Interest-only periods of six months and payment holidays also now available.
VAT cut from 20 to 5 per cent for firms in hospitality and tourism sectors is extended to March 2021.
Other firms who have deferred VAT bills under New Payment Scheme will be allowed to pay it off in 11 interest-free payments in 2021-22 financial year instead of one full payment in March 2021.
'Time to Pay' self-assessment tax system extended to January 2022.
Workers will be able to pick up 77 per cent of their wages for doing as little as a third of their regular hours under plans unveiled by Rishi Sunak today.
The Chancellor unveiled the Job Support Scheme (JSS) to 'directly support' wages of staff in 'viable' roles for six months from November.
It will replace the furlough, or Job Retention Scheme (JRS), which is due to come to an end on October 31.
But the new scheme is less generous than furlough, with the Chancellor admitting it was too expensive to continue after October.
Under the new system employees' wages receive a maximum subsidy of 22 per cent from the Treasury, depending on how many hours they work.
But firms must pick up the additional 55 per cent, compared to 20 per cent under furlough.
It has sparked fears that many businesses will simply lay staff off instead of taking it up.
Paul Johnson, director of the IFS think tank, said: 'The new job support scheme represents a significant new intervention from government to support jobs through the crisis.
'But it is significantly less generous than the furlough scheme it replaces – though remarkably the Chancellor provided no indication of the likely cost of the scheme.
'He is trying to plot a difficult path between supporting viable jobs while not keeping people in jobs that will not be there once we emerge from the crisis.
'With employers now having to pay at least 55 per cent of the normal wages of their employees it is clear that many jobs will be lost over the coming months.'
At a live press conference this afternoon, Mr Sunak said: 'We can't continue to provide the same degree of support as we did at the beginning of this crisis and sustain it at that level.
'It's not affordable and not sustainable for a prolonged period of time.'
The Chancellor said the new jobs support scheme was 'more targeted' than furlough, and would help businesses who really need assistance.
'Larger businesses on this scheme will have to demonstrate that their revenues are falling as a result of what is happening to them,' he explained.
He added: 'You can use this scheme in conjunction with the jobs retention bonus that we put in place in the summer, and if you look at the combined impact of those, I think it provides a very significant cash incentive for businesses to retain their staff even if that means on reduced hours.'
Under the JSS plans, businesses will continue to pay workers for the hours they work. But the JSS is designed to help support workers wages for the hours they lose due to restrictions on working hours or because of a lack of business due to economic decline.
For the regular hours that they are missing out on working, the Treasury and the employer will both pay a third of their wages.
This means workers will get all their wages for hours they do work, and two-thirds of their waged for hours they have not worked.
The Chancellor unveiled the Job Support Scheme (JSS) to 'directly support' wages of staff in 'viable' roles for six months from November
To be eligible, workers have to have worked at least a third (33 per cent) of their regular hours. In this scenario, a worker would receive 77 per cent of their total wages, with the firm picking up 55 per cent and the Treasury 22 per cent.
This is a large increase from the furlough scheme, under which firms paid in just 20 per cent of wages, with the Government picking up the rest.
The Government contribution will be capped at £697.92 a month and it will be paid in arrears – your employer pays you and then claims it back.
But the Treasury hopes it will allow businesses to reduce staff hours instead of simply laying them off because they cannot afford to pay them.
Businesses will not be able to issue redundancy notices to employees while taking part in the Jobs Support Scheme and there will be restrictions on capital distributions to shareholders.
But experts warned that the scheme still leaves a lot of pressure on businesses and may not be enough to see off a vast wave of redundancies when it replaces furlough on November 1.
An example of how much workers can expect
John is a builder earning £10 and working a 40-hour week. He takes home £400 before tax.
Under Rishi Sunak's scheme he would work a minimum of 13 hours a week. For those 13 hours he would be paid £308. Of that, his employer pays £220 and the Treasury £80.
The Treasury also gave an example of how it would work:
Beth normally works five days a week and earns £350 a week. Her company is suffering reduced sales due to coronavirus.
Rather than making Beth redundant, the company puts Beth on the Job Support Scheme, working two days a week (40 per cent of her usual hours).
Her employer pays Beth £140 for the two days she works.
And for the time she is not working (three days or 60 per cent, worth £210), she will also earn two-thirds, or £140, bringing her total earnings to £280, or 80 per cent of her normal wage.
The Government will give a grant worth £70 (a third of hours not worked, equivalent to 20 per cent of her normal wages) to Beth's employer to help them keep her job.
Source: HM Treasury
The scheme is geared up so that the more people work, the less Treasury has to pay. But that does risk firms deliberately cutting hours in order to obtain more of a grant.
Who is eligible
All employers with a UK bank account and UK PAYE schemes can claim the grant.
Neither the employer nor the employee needs to have previously used the Coronavirus Job Retention Scheme.
Large businesses will have to meet a financial assessment test, so the scheme is only available to those whose turnover is lower now than before experiencing difficulties from Covid-19.
There will be no financial assessment test for small and medium enterprises (SMEs).
A Treasury spokesman said: 'In order to support viable jobs, for the first three months of the scheme the employee must work at least 33 per cent of their usual hours.
'After three months, the Government will consider whether to increase this minimum hours threshold.'
Businesses will get more time to pay back emergency loans and tax bills under plans unveiled today by the Chancellor.
More than a million companies which took out Bounce Back Loans will be allowed to pay them back over 10 years instead of six under a 'pay as you grow' plan outlined by Rishi Sunak in the Commons.
He told MPs that this would almost halve the monthly repayments facing businesses, amid fears that economic damage may last monger than first thought.
It brings the Bounce back Loans scheme into line with Germany's Schnellkredit – which allows repayments over a decade but charges higher interest.
Firms will also be able to take advantage of interest-only repayments for up to six months and be able to take payment holidays for the same period of time, once they have made six repayments.
In addition the Chancellor has extended the availability of Bounce Back Loans, Coronavirus Business Interruption Loans, Coronavirus Large Business Interruption Loans and the Future Fund until November 30.
So far these have paid out more more than a million loans totalling around £58billion.
In addition a new payment scheme will give more breathing space for more than £30 billion of deferred VAT payments, allowing companies and VAT-registered sole traders to make 11 interest-free payments in 2021-22 rather than a lump sum at the end of March.
The temporary 15 per cent VAT cut for tourism and hospitality that took the rate from 20 per cent to 5 per cent will be extended until the end of March.
But not all firms believe it will work.
Jon Kay, director of Bury St Edmunds-based Camp Tails Doggy Daycare, said: 'Pay-as-you-grow, interest-only terms, additional payment holidays, it's ultimately putting lipstick on a pig. Debt is debt and for many small businesses struggling to bounce back in a brutal environment it will be the final nail in the coffin. The Chancellor needs to announce protection for employers as much as employees. The latter don't exist without the former, after all.'
Mixed reaction
Trade union and business leaders have spoken out in support the new scheme – some of it caveated.
TUC general secretary Frances O'Grady said: 'Unions have been pushing hard for continued jobs support for working people.
'We are pleased the Chancellor has listened and done the right thing. This scheme will provide a lifeline for many firms with a viable future beyond the pandemic.
'But there's still unfinished business. Unworked hours under the scheme must not be wasted.
'Ministers must work with business and unions to offer high-quality retraining, so workers are prepared for the future economy.'
Others praised the Chancellor's 'bold steps', saying hundreds of thousands of jobs will be saved.
Dame Carolyn Fairbairn, director general of the CBI, said it was right to target help on jobs with a future.
'These bold steps from the Treasury will save hundreds of thousands of viable jobs this winter,' she said.
'Wage support, tax deferrals and help for the self-employed will reduce the scarring effect of unnecessary job losses as the UK tackles the virus. Further business rates relief should remain on the table.
'The Chancellor has listened to evidence from business and unions, acting decisively. It is this spirit of agility and collaboration that will help make 2021 a year of growth and renewal.'
But economists were less bowled over.
Clare McNeil, associate director of the left-leaning IPPR think tank said: 'Choosing to continue to support the incomes of workers and businesses through the hugely uncertain next six months is the right decision. However, getting the design of the scheme right will be imperative to avoid layoffs, and the plan has three major flaws.'
She warned it may not go far enough in preventing lay-offs because firms will have to pay too much of wages, plus other costs like NI contributions and pensions.
The scheme is mitigated by the Job Retention Bonus, which pays out £1,000 for each worker kept on, but it ends in January and is closed to new applications
She also said the requirement for at least a third of hours to be worked would badly affect firms unable to trade at all.
She added: 'However the scheme is more generous than expected in allowing any SME to be eligible for support, while large businesses have to demonstrate that their turnover has been hit but Coronavirus.'
The scheme has echoes of the German Kurzarbeit job subsidy measure, which has been recently extended until the end of 2021.
It allows employers to reduce employees' hours while keeping them in a job. The Government pays workers a percentage of the money they would have got for working those lost hours.
For example, for someone who usually works 37 hours a week but can only now work 17, a company could pay them their full wage for the 17 hours but it could charge the Government to pay a proportion of the remaining 20 hours.
According to the Munich-based Institute for Economic Research, at the height of the pandemic, half of all German firms had at least some of their staff on the scheme.
Influential British political figures including former prime minister Gordon Brown urged the Government to bring in such a scheme, or a similar French-style system, after the furlough scheme ends in October.
More help for the self-employed but many miss out
Chancellor Rishi Sunak said that the Government will cover 20 per cent of the earnings of self-employed people after the current scheme, which covers 80 per cent of earnings, runs out.
More than a million people will have to fend for themselves after the Chancellor failed to plug holes in his support for the self-employed, a trade body has warned.
He also promised to defer income tax self-assessment, a measure that will be 'of particular importance' to self employed people.
But the help misses out around 1.5 million people who have been given far too little support through the crisis, said Andy Chamberlain, the director of policy at the Association of Independent Professionals and the Self-Employed (IPSE).
Only sole traders who filed tax returns in the financial year ending April 2019 were eligible for the previous Self-Employment Income Support Scheme (Seiss), and the Chancellor did not appear to close these gaps on Thursday.
'The support for the self-employed announced today is woefully inadequate.
'Although it is right for the Chancellor to extend Seiss, the support announced today still excludes one in three self-employed people,' Mr Chamberlain said.
'Limited company freelancers and the newly self-employed almost entirely missed out on support in the last lockdown and have faced bleak months of financial devastation.
'Now they face a dark winter ahead unless the government does more for them.'
He added that the 20 per cent cap on support is likely to prove insufficient for many.
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Limelight / Music
Limelight / Reviews
Mac Miller's posthumous "Circles" sets a high bar for 2020
mac miller review design
Collin Maguire
Mac Miller's posthumous release "Circles" acts as a companion to Miller's 2018 release "Swimming." However, the two projects contain different styles. It's apparent Miller poured his heart into this one.
Miller died of an accidental drug overdose in September 2018, creating ripples in the realm of hip-hop and the music industry as a whole. Miller and his team were in the process of creating the project at the time of his passing, but his vision was never lost.
His family left a post on his Instagram stating, "This is a complicated process that has no right answer. No clear path. We simply know that it was important to Malcolm for the world to hear it."
"Circles" carries a different sound in comparison to the rest of Miller's discography. Miller burst into the scene as a charismatic rapper from Pittsburgh and began to gain some traction with his mixtape "K.I.D.S." This tape was home to some fan favorites like "Nikes On My Feet" and "Knock Knock."
With this final release, Miller's career has finally come full circle. As he became a larger entity in hip-hop, his own artistic flare began to develop. As Miller grew, he began to write about topics many artists would struggle to put to paper. In today's music lens, it's clear he always possessed a strong share of versatility.
Living a free-spirited life was always something Miller preached to his fans. Despite his demons, his smile remained true, and he led a lot of his listeners out of darkness. "Circles" was the perfect bit of closure for him and his fanbase.
Miller worked closely with producer Jon Brion throughout the process of "Swimming." Brion acted as the figure to carry out Miller's vision with "Circles." Brion has worked with artists of the likes of Missy Elliott, Kanye West, Beyoncé and many more.
The slow-paced title track gets the ball rolling on the album. In this track, we hear Miller sing about being stuck in cycles everyday of his life. This sets the stage for him to tackle subjects surrounding his cluttered mind.
Throughout his career, Miller blossomed in terms of lyricism, but his singing voice wasn't fully established until this final turn of his career. "Circles" is the perfect intro track as Miller fills the room—or headphones—with his voice.
"Blue World" is an immediately gripping track with a cool little sample of The Four Freshmen's,"It's a Blue World." But a trippy sample switch-up, courtesy of Disclosure's Guy Lawrence, ties everything together as the first hook ensues.
As the most hip-hop oriented song on the track, Miller's artistry never seemed blurred. Themes regarding the ups and downs of life were heavily involved. Miller reminisces about the good times listening to music with his ex-girlfriend, pop star Ariana Grande.
This track displays a high production level and how Miller could really walk hand-in-hand with any sound his team threw at him.
"Good News" was dropped as the standalone single for the album, accompanied by the announcement of this final album's release.
In this track, we hear Miller alluding to depression, a common topic in his last release, "Swimming." The song begins with Miller following a light instrumental by saying, "I spent the whole day in my head."
Miller was always a free-spirited character, and in this track, some of that energy seeps right through. Another line in the first verse reads, "Got the cards in my hand, I hate dealing, yeah," alluding to Miller wishing he could just live a simple life rather than being an important figure and meaning so much to others.
Miller excelled vocally and lyrically in this track regarding past relationships. He sings on a delicate beat produced by Eric Dan and DAVID x ELI.
This track is among the most unique tracks on the record. It's clear the producers took working with Miller very seriously, as "Woods" makes up one of the many beautiful songs on "Circles."
"Surf" was produced by Miller and Brion and acts as a perspective into the artist's mental state. Miller discusses how he views himself and the self-esteem issues that clouded his judgement each day.
Miller's laid-back yet appealing singing voice compliments some beautiful sections filled with guitar playing. "Surf" stands out as one of the top performances on the project and acts as a great lead into the concluding track "Once A Day."
This track was originally leaked in October at the "Mac Miller: A Celebration of Life" concert. The song acts as a wonderful conclusion and is the shortest track on "Circles," less than three minutes in duration.
Miller begins the final verse by saying, "Don't keep it all in your head," again, another allusion to mental health struggles. But this reference acts as a direct message to the listener, encouraging them to open up and talk about their struggles.
Malcom McCormick displayed his dynamic talent through personal introspection in his music, and "Circles" was the perfect way to close out the legacy. Thank you for all you've done; rest up.
Final Verdict: 9/10
Collin Maguire, Author
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Q: Divisibility of a 7-digit number by 21 How many 7-digit numbers formed by using only the digits $3$ and $7$, and divisible by $21$ are there?
Well, I let the number be abcdefg and:

*

*Placing $3$ at any position gives $0 \bmod 3$

*Placing $7$ at all positions a,b,c,d,e,f,g gives $1 \bmod 3$

Then for abcdefg to be divisible by $3$, there would be $7C3 + 7C0 +7C6$ options
Next:

*

*Placing $7$ at any position gives $0 \bmod 7$

*Placing $3$ at $a$ gives $3 \bmod 7$, $b$ gives $2 \bmod 7$, $c$ gives $6 \bmod 7$, $d$ gives $4 \bmod 7$, $e$ gives $5 \bmod 7$, $f$ gives $1 \bmod 7$, $g$ gives $3 \bmod 7$

And then I was going to do the same thing as before but how do I sort of link it back to divisibility by $21$? Is it PIE or something? Any hint would be appreciated. Thank you.
Edit: Oops, I realised that I've solved it already. So I used the mod values for 7 obtained above and selected any 3 or 6 out of a,b,c,d,e,f,g , to take the value 3, that would add up to be divisible by 7 and since the sum of the digits would add up to 3, they would be divisible by 21 naturally. The combinations are (a,f,g), (a,c,e), (c,e,g), (b,d,f), (a,b,c,d,e,f) and (b,c,d,e,f,g). Therefore, the answer is 6 which is correct. But if anyone has any other solution that is simpler or more elegant, do let me know as well!

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Elegantly dressed in a beautiful blue evening gown, 24-year old Christine Barbier smiled and walked gracefully, waving to the cheering crowd at the International Conference Centre of Seychelles, minutes after she was crowned as the country's new beauty queen last night.
The new Miss Seychelles...another world, who hails from the district of Cascade, also went home with three other titles. She won the Miss Sports, Miss Talent and Miss Personality titles. Miss Barbier is an accounts assistant and her wish is to one day become a successful businesswoman.
Lianne Barbe who comes from the island of Praslin, was crowned first princess and also Miss Photogenic. The 19-year old also won Miss People's choice title.
Samiya Albert, is this year's second princess. The 18-year old who is from Pointe Larue wore a long glittering ivory-coloured evening gown, which the judges voted as the best evening wear.
Sophie Rosette, who was among the six finalists was the contestant who won the nest.effort award for her beauty with a purpose project.
The two other contestants who were in the final were Samira Nicoire and Rita-Maria Durup.
Among those present at the event were the Minister for Tourism and Culture Alain St.Ange, the principal secretaries of the Ministry of Tourism and Culture Anne Lafortune and Benjamine Rose, and the chief executive of the Seychelles Tourism Board.
Other girls who took part in the pageant this year were Hemma Hoffman, Anais Nourice, Juanita Porice, Camilla Spivey, Ruth Bonte and Kimberley Anacoura.
Members of the jury were Miss Seychelles 2006 Wendy Boniface, chief executive of the Seychelles Trading Company Veronique Laporte, tourism entrepreneur Angie Williams-Leiss, the chairperson of the Round Table Seychelles Abison de Giorgio, the chief executive of Noussira enterprise Noussira Daman, director general in the Ministry of Foreign Affairs and Transport as well as the managing director of Maven, Kosta Todorovic.
During the pageant, the girls paraded in their casual and beach wears as well as their evening gowns before the six finalists were chosen.
The six contestants then had to answer questions from members of the jury. Before the new beauty queen and her princesses were crowned, the 2015 Miss Seychelles...another world Linne Freminot paraded on the catwalk.
Top local artists such as Angie, who was also one of the contestants last year, Leo, Joseph Sinon and Antoinette Dodin performed during the event.


text:
Муку́нда:
 Мукунда — одно из имён Кришны в индуизме, в переводе означающее «дарующий освобождение».
 Мукунда Госвами (в миру Майкл Грант; род. 1942) — кришнаитский гуру, музыкант, публицист и писатель.
 Мукунда Даса — кришнаитский святой XVI века.
 Мукунда Датта — кришнаитский святой  XV—XVI века.

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction and Preliminaries}

In~\cite{MZ} authors consider the distributive lattices of all order ideals
of the fence poset and crown poset ordered by inclusion, and they are able to prove
recursive formulas for their Whitney numbers.
In this paper, using purely combinatorial methods, we solve these recursions giving
explicit closed formulas for the corresponding rank polynomials.
Moreover, in \S~\ref{S:as} we consider a more general class of fence posets, namely
fences with higher asymmetric peaks, and we get explicit closed formulas for Whitney
numbers of lattices of their order ideals.

For others combinatorial results about lattices of order ideals of finite posets
and their Whitney numbers, we remind to~\cite{CGW,Gri,Ste1,Ste2}.

In the sequel we collect some definitions, notations and results that will
be used in the following.
For $x \in \mathbb{R}$ we let $\fl{x}=\max\{n \in \mathbb{Z} : n \leq x\}$ and
$\rf{x}=\min\{n \in \mathbb{Z} : n \geq x\}$; for any $n,m \in \mathbb{N}$, $n \leq m$, we
let $\[n,m\]=\{t \in \mathbb{N} : n \leq t \leq m\}$, and $\[n\]=\[1,n\]$, therefore
$\[0\]= \emptyset$. For any complex number $a$, we define the \emph{rising
factorial} as $\(a\)_{0}=1$ and $\(a\)_{m}=\prod_{j=0}^{m-1}\(a+j\)$ for any
$m \in \mathbb{N} \setminus \{0\}$.The cardinality of a set ${\mathcal X}$ will be denoted by
$\# {\mathcal X}$.

We follow~\cite{And,GrKl,Sta2} for combinatorics notations and
terminology.
We recall that a \emph{ranked poset} is a poset $P$ with a function
$\rho:P \longrightarrow \mathbb{N}$, called rank, such that $\rho\(y\)=\rho\(z\)+1$
whenever $z$ is covered by $y$ in $P$ and $\min\{\rho\(z\) : z \in P\}=0$.
The \emph{rank polynomial} of a ranked finite poset $P$ is the polynomial
$$\sum_{z \in P}X^{\rho\(z\)}=\sum_{j \geq 0} \omega_{j} X^{j},$$
where $\omega_{j}=\# \{z \in P : \rho\(z\)=j\}$ are called
\emph{Whitney numbers} of $P$.

An \emph{order ideal} of a poset $P$ is a subset $I \subset P$ such that if
$y \in I$ and $z \leq y$, then $z \in I$; it is well known that the set of
all order ideals of $P$ ordered by inclusion is closed under unions and
intersections, and hence forms a distributive lattice: we denote it by
${\mathcal J}\(P\)$, viz. ${\mathcal J}\(P\)=\{I \subset P : I \text{ is an order ideal}\}$.
It is not hard to see that its rank function is the cardinality of order
ideals.

Given a finite poset $\(P,\leq\)$, we denote with $W_{P}\(k\)$ the
$k$--th Whitney numbers of the ranked poset of all order ideals of $P$, i.e.
$W_{P}\(k\)=\# \{I \in {\mathcal J}\(P\) : \rho\(I\)=j\}$, where $\rho$ is the rank
function of ${\mathcal J}\(P\)$, and the rank polynomial of ${\mathcal J}\(P\)$ is denoted
by ${\mathcal R}_{P}\(X\)$, i. e. ${\mathcal R}_{P}\(X\)=\sum_{k \geq 0} W_{P}\(k\) X^{k}$.

We denote by ${\mathcal Z}_{n}$ the \emph{fence} poset of order $n$, viz. the poset
$\{z_{1}, \ldots ,z_{n}\}$ in which
$z_{2j-1} \triangleleft z_{2j} \triangleright z_{2j+1}$, for all
$j \geq 1$, are the cover relations, by ${\mathcal I}_{n}\(k\)$ the set of order
ideals of ${\mathcal Z}_{n}$ with cardinality $k$, and by $f_{n,k}$ the Whitney
numbers of the poset of all order ideals of a fence of order $n$, viz.
$f_{n,k} = \# {\mathcal I}_{n}\(k\)=W_{{\mathcal Z}_{n}}\(k\)$. \\
We denote by ${\mathcal Y}_{n}$ the \emph{crown} poset of order $2n$, viz. the poset
$\{\zeta_{0}, \ldots ,\zeta_{2n-1}\}$ in which the cover relations are
the following: for any $h \in \{0, \ldots n-1\}$ and $k \in \[n\]$,
$\zeta_{2h} \triangleleft \zeta_{2k-1}$ if and only if
$\left|2h-2k+1\right| \equiv 1 \pmod{2n}$, therefore
$$\rho\(\zeta_{j}\)=\begin{cases}
0 & \text{ if } j \equiv 0 \pmod{2}, \\
1 & \text{ if } j \equiv 1 \pmod{2}.
\end{cases}$$
We also denote by ${\mathcal O}_{n}\(k\)$ the set of order ideals of ${\mathcal Y}_{n}$ with
cardinality $k$, and by $c_{n,k}$ the Whitney numbers of the poset of all
order ideals of a crown of order $2n$, viz.
$c_{n,k} = \# {\mathcal O}_{n}\(k\)=W_{{\mathcal Y}_{n}}\(k\)$.

Finally we recall, gluing together, Propositions $1$, $3$ and $5$ of~\cite{MZ},
which give recursions for the sequences $f_{n,k}$ and $c_{n,k}$.

\begin{prop} \label{P:5}
For any integer $n$ the recurrence identity
$$\begin{cases}
f_{2n,k} & = f_{2n-1,k} + f_{2n-2,k-2} \\
f_{2n+1,k} & = f_{2n,k-1} + f_{2n-1,k}
\end{cases}$$
holds, where
$$\begin{cases}
f_{n,k} = 0 & \qquad \text{ if } \, k \notin \[0,n\] \; \text{ or } \,
                       n < 0 \\
f_{n,0} = 1 & \qquad \text{ for all } \, n \in \mathbb{N}
\end{cases}$$
are the initial values.

Moreover, with the same initial values the formula
$$f_{n+4,k+2}=f_{n+2,k+2}+f_{n+2,k+1}+f_{n+2,k}-f_{n,k}$$
holds, for all $0 \leq k \leq n \in \mathbb{N}$.

Furthermore,
\begin{eqnarray*}
c_{n+2,k+3} & = & f_{2n+3,k+3} + f_{2n+1,2n+1-k} \\
c_{n+2,k+2} & = & c_{2n+1,k} + f_{2n+3,k+2} - f_{2n-1,k} \\
c_{n+2,k+2} & = & f_{2n+4,k+2} - f_{2n,k}
\end{eqnarray*}
hold, for all $n \in \mathbb{N}$ and all $0 \leq k \leq 2n$.
\end{prop}

\section{Closed Formulas for Whitney Numbers}

\label{S:cf}

We need the following Proposition, whose proof can be found
in~\cite{Sta2}.

\begin{prop} \label{P:s}
For all non--negative integers $k \leq n$,
\begin{eqnarray*}
\# \{\mathbf{\overline{x}}=\(x_{1}, \ldots ,x_{k}\) \in
   \(\mathbb{N} \setminus \{0\}\)^{k} : \sum_{j=1}^{k} x_{j} = n\} & = &
   \binom{n-1}{k-1}, \\
\# \{\mathbf{\overline{x}}=\(x_{1}, \ldots ,x_{k}\) \in \mathbb{N}^{k} :
   \sum_{j=1}^{k} x_{j} = n\} & = & \binom{n+k-1}{k-1}, \\
\# \{\mathbf{\overline{x}}=\(x_{1}, \ldots ,x_{k}\) \in \mathbb{N}^{k} :
   \sum_{j=1}^{k} x_{j} \leq n\} & = & \binom{n+k}{k},
\end{eqnarray*}
hold.
\end{prop}

\begin{thm} \label{T:main}
For all $k,v \in \mathbb{N}$ such that $k \leq 2v+1$,
$$f_{2v+1,k} = \binom{v+1}{k} + \sum_{j \geq 1}
\frac{\(k-2j+1\)_{j-1}\(v+j-k+2\)_{k-2j}}{j!\(k-2j-1\)!}$$
holds.
\end{thm}
\begin{proof}
For all integers $0 \leq k \leq n$, we can write
$$f_{n,k} = \# {\mathcal I}_{n}\(k\) = \sum_{j \geq 0} {\mathcal A}\(n,k,j\) =
\sum_{j=0}^{\min\{\fl{\frac{n}{2}},\rf{\frac{k}{2}}-1\}} {\mathcal A}\(n,k,j\),$$
where
$${\mathcal A}\(n,k,j\)=\# \{{\mathcal J} \in {\mathcal I}_{n}\(k\) :
\# \{x \in {\mathcal J} : \rho\(x\)=1\}=j\};$$
thus we have that $f_{n,0}=1$,
$f_{n,1} = \# \{x \in {\mathcal Z}_{n} : \rho\(x\)=0\} = \fl{\frac{n+1}{2}} =
\rf{\frac{n}{2}}$, and ${\mathcal A}\(n,k,0\) = \binom{f_{n,1}}{k}$.

Consider a fence ${\mathcal Z}_{n}$ with odd cardinality, i.e. $n=2v+1$
for some $v \in \mathbb{N}$, and write it as the poset
$\{z_{1}, \ldots ,z_{2v+1}\}$ in which
$z_{2\alpha-1} \triangleleft z_{2\alpha} \triangleright z_{2\alpha+1}$,
for all $\alpha \geq 1$, are the cover relations. \\
For any given ${\mathcal J} \in {\mathcal I}_{2v+1}\(k\)$ (with $k \geq 2$) such that
$\# \{x \in {\mathcal J} : \rho\(x\)=1\}=j \geq 1$ we can split the set
$\{x \in {\mathcal J} : \rho\(x\)=1\}$ in $r$ separated non-empty blocks
${\mathcal X}_{1}, \ldots ,{\mathcal X}_{r}$, such that $\sum_{t=1}^{r} \# {\mathcal X}_{t} =j$;
$z_{2a},z_{2b} \in {\mathcal J}$ with $1 \leq a < b \leq v$ are in the same block
if and only if $z_{2c} \in {\mathcal J}$ for all $c$ such that $1 \leq a < c < b \leq v$.
Each ${\mathcal X}_{t}$ determines $2 \# {\mathcal X}_{t} +1$ elements in ${\mathcal J}$, so this
decomposition fix $\sum_{t=1}^{r} \(2 \# {\mathcal X}_{t} +1\) = 2j+r$
elements of ${\mathcal J}$ ($j$ of these have rank $1$, and the others $j+r$ have rank
$0$), and obviously the others can be chosen in
$\binom{v+1-\(j+r\)}{k-\(2j+r\)}$ ways between the remainder elements with
rank $0$. \\
Moreover, the number of such decompositions ${\mathcal X}_{1}, \ldots ,{\mathcal X}_{r}$ is
$\# {\mathcal C}\(j,r\)$ times the the total numbers of shifts of all blocks
${\mathcal X}_{1}, \ldots ,{\mathcal X}_{r}$, which can be evaluated in the following way:
at least one element of rank $1$ has to be into the slot between the blocks
${\mathcal X}_{t}$ and ${\mathcal X}_{t+1}$, for any $t \in \[r-1\]$, and the others
$v-\(j+r-1\)$ elements can be freely distributed into the $r+1$ slots, viz.
before ${\mathcal X}_{1}$, between ${\mathcal X}_{t}$ and ${\mathcal X}_{t+1}$, for any
$t \in \[r-1\]$, and after ${\mathcal X}_{r}$, thus from Proposition~\ref{P:s}
$\binom{v-j+1}{r}$ is the searched value.

\noindent
Therefore if we define ${\mathcal C}\(\mu,\nu\)=
\{\mathbf{\overline{x}}=\(x_{1}, \ldots ,x_{\nu}\) \in
   \(\mathbb{N} \setminus \{0\}\)^{\nu} : \sum_{j=1}^{\nu} x_{j} = \mu\}$
for any $1 \leq \nu \leq \mu$, from Proposition~\ref{P:s} we have
$\# {\mathcal C}\(\mu,\nu\) = \binom{\mu-1}{\nu-1}$, hence for any $j \geq 1$
\begin{eqnarray*}
{\mathcal A}\(2v+1,k,j\) & = & \sum_{r=1}^{j}
\sum_{\substack{\mathbf{\overline{x}} \in {\mathcal C}\(j,r\) \\ 2j+r \leq k \\
j+r-1 \leq v}} \binom{v-j+1}{r} \binom{v+1-\(j+r\)}{k-\(2j+r\)} \\
& = & \sum_{r=1}^{j}
\sum_{\mathbf{\overline{x}} \in {\mathcal C}\(j,r\)} \binom{v-j+1}{r}
\binom{v+1-\(j+r\)}{k-\(2j+r\)} \\
& = & \sum_{r=1}^{j} \binom{j-1}{r-1} \binom{v-j+1}{r}
\binom{v+1-\(j+r\)}{k-\(2j+r\)}.
\end{eqnarray*}

Therefore we have
\begin{eqnarray*}
\lefteqn{f_{2v+1,k} = \binom{v+1}{k} +
\sum_{j=1}^{\min\{v,\rf{\frac{k}{2}}-1\}} \sum_{r=1}^{j} \binom{j-1}{r-1}
\binom{v-j+1}{r} \binom{v+1-\(j+r\)}{k-\(2j+r\)}} \\
& & = \binom{v+1}{k} + \sum_{j \geq 1}
\sum_{r \geq 1} \binom{j-1}{j-r} \binom{v-j+1}{r}
\binom{v+1-\(j+r\)}{k-\(2j+r\)}
\end{eqnarray*}

Writing the sum over $r$ in hypergeometric notation and applying
Chu--Vandermonde summation, see~\cite{Fin,GR,Krat}, we get
\begin{flalign*}
& \sum_{r \geq 1} \binom{j-1}{j-r} \binom{v-j+1}{r}
\binom{v+1-\(j+r\)}{k-\(2j+r\)} \\
& = \frac{{}_{2}F_{1}\[\begin{matrix} 1-j,1+2j-k\\ 2 \end{matrix};1\]
\(v+j-k+2\)_{k-2j}}{\(k-2j-1\)!} \\
& = \frac{\(k-2j+1\)_{j-1}\(v+j-k+2\)_{k-2j}}{j!\(k-2j-1\)!},
\end{flalign*}
and the desired result follows.
\end{proof}

\begin{cor}
For any $v \in \mathbb{N}$ and all $0 \leq k \leq 2v+1$, the sequence
$f_{2v+1,k}$ is increasing in $v$, viz. $f_{2\(v+1\)+1,k} > f_{2v+1,k}$.
\end{cor}
\qed

\begin{defn} \label{D:star}
Let $\(P_{1}, \leq_{1}\)$, $\(P_{2}, \leq_{2}\)$ be finite posets with
cover relations $\triangleleft_{1}$ and $\triangleleft_{2}$, respectively,
and let $x_{1} \in P_{1}$, $x_{2} \in P_{2}$ be minimal elements. \\
We consider a new element $\widetilde{x}$ which does not belong
to $P_{1} \biguplus P_{2}$ and we define a new poset
$\(P_{1}\(x_{1}\) \circledast P_{2}\(x_{2}\), \leq\)$ with
cover relations $\triangleleft$, where
$$P_{1}\(x_{1}\) \circledast P_{2}\(x_{2}\)=
P_{1} \biguplus P_{2} \biguplus \{\widetilde{x}\},$$
and for any $x,y \in P_{1}\(x_{1}\) \circledast P_{2}\(x_{2}\)$ we have
$x \triangleleft y$ if and only if one of the following conditions holds:
\begin{description}
\item{$\circ$} $x,y \in P_{1}$ and $x \triangleleft_{1} y$ in $P_{1}$,
\item{$\circ$} $x,y \in P_{2}$ and $x \triangleleft_{2} y$ in $P_{2}$,
\item{$\circ$} $x_{1} \triangleleft \widetilde{x}$,
\item{$\circ$} $x_{2} \triangleleft \widetilde{x}$.
\end{description}
\end{defn}

\begin{thm} \label{T:1fl}
Let $\(P_{1}, \leq_{1}\)$, $\(P_{2}, \leq_{2}\)$ be finite posets,
$x_{1} \in P_{1}$, $x_{2} \in P_{2}$ be minimal elements, and
$\widehat{P}=P_{1}\(x_{1}\) \circledast P_{2}\(x_{2}\)$;
then
$${\mathcal R}_{\widehat{P}}\(X\) = {\mathcal R}_{P_{1}}\(X\) {\mathcal R}_{P_{2}}\(X\)
 + X^{3} {\mathcal R}_{P_{1} \setminus \{x_{1}\}}\(X\)
{\mathcal R}_{P_{2} \setminus \{x_{2}\}}\(X\)$$
holds.
\end{thm}
\begin{proof}
Let us write ${\mathcal J}\(\widehat{P}\)= \biguplus_{k=0}^{\# \widehat{P}}
{\mathcal J}_{k}$, where ${\mathcal J}_{k}=\{I \in {\mathcal J}\(\widehat{P}\) : \rho\(I\)=\# I =k\}$,
thus
$$W_{\widehat{P}}\(k\)=\# {\mathcal J}_{k}=
\# \{I \in {\mathcal J}_{k} : \widetilde{x} \notin I\} +
\# \{I \in {\mathcal J}_{k} : \widetilde{x} \in I\}.$$
It is not hard to see that
$$\# \{I \in {\mathcal J}_{k} : \widetilde{x} \notin I\}=
\sum_{j=0}^{k} W_{P_{1}}\(j\) \cdot W_{P_{2}}\(k-j\)$$
and
\begin{eqnarray*}
\# \{I \in {\mathcal J}_{k} : \widetilde{x} \in I\} & = &
\# \{I \in {\mathcal J}_{k} : x_{1},x_{2},\widetilde{x} \in I\} \\
& = & \sum_{j=0}^{k-3} W_{P_{1} \setminus \{x_{1}\}}\(j\) \cdot
W_{P_{2} \setminus \{x_{2}\}}\(k-3-j\),
\end{eqnarray*}
and the desired result follows.
\end{proof}

\begin{thm} \label{T:even}
For all $k,v \in \mathbb{N}$ such that $k \leq 2v$,
\begin{eqnarray*}
f_{2v,k} & = & \sum_{j \geq 0} \sum_{r \geq 0}
\binom{j}{r} \binom{v-j}{r} \binom{v-\(j+r\)}{k-\(2j+r\)} \\
& = & \sum_{j \geq 0}
\frac{\(k-2j+1\)_{j}\(v+j-k+1\)_{k-2j}}{j!\(k-2j\)!}
\end{eqnarray*}
holds.
\end{thm}
\begin{proof}
For any $n \in \mathbb{N} \setminus\{0\}$ write the fence poset
${\mathcal Z}_{n}$ as the poset $\{z_{1}, \ldots ,z_{n}\}$ in which
$z_{2j-1} \triangleleft z_{2j} \triangleright z_{2j+1}$, for all
$j \geq 1$, are the cover relations, so $\rho\(z_{j}\)=0$ if and
only if $j \equiv 1 \pmod{2}$ and $\rho\(z_{j}\)=1$ if and only
if $j \equiv 0 \pmod{2}$.
If we consider $P_{1}={\mathcal Z}_{2v+1}=\{a_{1}, \ldots ,a_{2v+1}\}$,
$P_{2}={\mathcal Z}_{1}=\{b_{1}\}$, we have that
$P_{1}\(a_{2v+1}\) \circledast P_{2}\(b_{1}\) \simeq {\mathcal Z}_{2v+3}$,
and the desired result follows applying Theorems~\ref{T:1fl}
and~\ref{T:main}, and Chu--Vandermonde summation for hypergeometric
series as in the proof of Theorem~\ref{T:main}.
\end{proof}

\begin{cor}
For any $v \in \mathbb{N}$ and all $0 \leq k \leq 2v$, the sequence
$f_{2v,k}$ is increasing in $v$, viz. $f_{2\(v+1\),k} > f_{2v,k}$.
\end{cor}
\qed

From Proposition~\ref{P:5} and Theorem~\ref{T:even} we
immediately get the following result.

\begin{thm} \label{T:cr}
For all $k,n \in \mathbb{N}$ such that $k \leq 2n$,
\begin{eqnarray*}
c_{n,k} & = & \sum_{j \geq 0}
\frac{\(k-2j+1\)_{j-2}
\(n+j-k+1\)_{k-2j-2}}{j!\(k-2j\)!} \\
& \cdot & \[\(k-j-1\)_{2}\(n-j-1\)_{2}
-\(\(k-2j-1\)_{2}\)^{2}\]
\end{eqnarray*}
holds.
\end{thm}
\qed

Therefore Theorems~\ref{T:main}, \ref{T:even} and~\ref{T:cr} give the
solution of the recursive identities in Proposition~\ref{P:5}.

\section{Generalized Fences with higher asymmetric peaks}

\label{S:as}

Now we define an \emph{asymmetric peak} poset with two positive integers
parameters $\mu,\nu$.

\begin{defn}
Let $\mu,\nu \in \mathbb{N} \setminus\{0\}$; we define the poset
\emph{asymmetric peak} $\(\operatorname{AP}_{\mu,\nu}, \leq\)$
in the following way:
$\operatorname{AP}_{\mu,\nu} =
\{a_{j} : j \in \[\mu\]\} \biguplus \{b_{j} : j \in \[\nu\]\}
\biguplus \{\omega\}$, and the cover relations are
\begin{description}
\item{$\circ$} $a_{j} \triangleleft a_{j+1}$ for all $j \in \[\mu-1\]$,
\item{$\circ$} $b_{j} \triangleleft b_{j+1}$ for all $j \in \[\nu-1\]$,
\item{$\circ$} $a_{\mu} \triangleleft \omega$,
\item{$\circ$} $b_{\nu} \triangleleft \omega$.
\end{description}
\end{defn}

\begin{prop} \label{P:ap}
Let $\mu,\nu \in \mathbb{N} \setminus\{0\}$; then
$$W_{\operatorname{AP}_{\mu,\nu}}\(k\)= \begin{cases}
1 \qquad & \text{ if } k=0 \text{ or } k=\mu + \nu + 1 \\
k+1 & \text{ if } k \leq \min\{\mu,\nu\} \\
\min\{\mu,\nu\} + 1
    & \text{ if } \min\{\mu,\nu\} \leq k \leq \max\{\mu,\nu\} \\
1 + \mu + \nu - k & \text{ if } \max\{\mu,\nu\} \leq k
\end{cases}$$
holds, for any $k=0, \ldots ,
\# \operatorname{AP}_{\mu,\nu} = \mu + \nu + 1$.
\end{prop}
\begin{proof}
The result is clear is $k=0$ or $k=\mu + \nu + 1$.

We consider the case $\mu \leq \nu$, the case $\mu \geq \nu$ is
completely symmetric. \\
If $k \in \[\mu+\nu\]$ then any
$I \in {\mathcal J}\(\operatorname{AP}_{\mu,\nu}\)$ with $\rho\(I\)=\# I = k$
has the shape
$I=\{a_{j} : j \in \[r\]\} \biguplus \{b_{j} : j \in \[t\]\}$ with
\begin{equation} \label{E:tr}
r+t=k,
\end{equation}
so $W_{\operatorname{AP}_{\mu,\nu}}\(k\)$ equals
the number of solutions of~\eqref{E:tr} with the constraints
\begin{align*}
0 \leq r \leq k  \; \text{ and } \; 0 \leq t \leq k \qquad
   & \text{ if } k \leq \mu, \\
0 \leq r \leq \mu  \; \text{ and } \; k - \mu \leq t \leq k \qquad
   & \text{ if } \mu \leq k \leq \nu, \\
 k - \nu \leq r \leq \mu  \; \text{ and } \; k - \mu \leq t \leq \nu
   \qquad & \text{ if } \nu \leq k.
\end{align*}

The desired result follows.
\end{proof}

Results proved in \S~\ref{S:cf} allows to get explicit closed formulas
for Whitney numbers of lattices of order ideals of ``fences with
higher asymmetric peaks'', i.e. the alternate composition of fences and
asymmetric peaks by the operator $\circledast$,
see Definition~\ref{D:star}.

For example, we can consider a fence with one higher asymmetric peak, which
can be formally defined as the following poset $\(FAP\(w,x,y,z\),\leq\)$
where with $w,x,y,z \in \mathbb{N}$ and $w \equiv 1 \pmod{2}$:
$$\operatorname{FAP}\(w,x,y,z\)=\{a_{1}, \ldots ,a_{w},b_{1}, \ldots ,b_{x},
\omega, c_{1}, \ldots ,c_{y},d_{1}, \ldots ,d_{z}\},$$
where the cover relations are
\begin{description}
\item{$\circ$} $a_{2j-1} \triangleleft a_{2j} \triangleright a_{2j+1}$,
               for all $j \geq 1$,
\item{$\circ$} $a_{w} \triangleleft b_{1}$,
\item{$\circ$} $b_{j} \triangleleft b_{j+1}$ for all $j \in \[x-1\]$,
\item{$\circ$} $c_{j} \triangleleft c_{j+1}$ for all $j \in \[y-1\]$,
\item{$\circ$} $b_{x} \triangleleft \omega$,
\item{$\circ$} $c_{y} \triangleleft \omega$,
\item{$\circ$} $d_{1} \triangleleft c_{1}$,
\item{$\circ$} $d_{2j-1} \triangleleft d_{2j} \triangleright d_{2j+1}$,
               for all $j \geq 1$.
\end{description}
In Figure~\ref{F:as} the Hasse diagram of $\operatorname{FAP}\(7,10,6,7\)$ is depicted.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=12cm]{peak.eps}
\caption{$\operatorname{FAP}\(7,10,6,7\)$}
\label{F:as}
\end{center}
\end{figure}

Inside $\operatorname{FAP}\(w,x,y,z\)$ consider the subposets
\begin{eqnarray*}
{\mathcal P}_{1} & = & \{a_{1}, \ldots ,a_{w-2}\} \simeq {\mathcal Z}_{w-2}, \\
{\mathcal P}_{2} & = & \{a_{w},b_{1}, \ldots ,b_{x}, \omega, c_{1},
        \ldots ,c_{y}, d_{1}\} \simeq \operatorname{AP}_{x+1,y+1}, \\
{\mathcal P}_{3} & = & \{d_{3}, \ldots ,d_{z}\} \simeq {\mathcal Z}_{z-2};
\end{eqnarray*}
therefore
$\operatorname{FAP}\(w,x,y,z\)={\mathcal P}_{1} \biguplus
\{a_{w-1}\} \biguplus {\mathcal P}_{2} \biguplus \{d_{2}\} \biguplus {\mathcal P}_{3}$.

We have that
\begin{eqnarray*}
\operatorname{FAP}\(w,x,y,z\) & \simeq &
\({\mathcal P}_{1}\(a_{w-2}\) \circledast {\mathcal P}_{2}\(a_{w}\)\)\(d_{1}\)
\circledast {\mathcal P}_{3}\(d_{3}\) \\
& \simeq & {\mathcal P}_{1}\(a_{w-2}\) \circledast
\({\mathcal P}_{2}\(d_{1}\) \circledast {\mathcal P}_{3}\(d_{3}\)\)\(a_{w}\),
\end{eqnarray*}
therefore from Theorems~\ref{T:main}, \ref{T:1fl},
\ref{T:even}, \ref{T:cr} and Proposition~\ref{P:ap} we get an
explicit closed formulas for the rank polynomial of the distributive
lattice of all order ideals of the poset
$\operatorname{FAP}\(w,x,y,z\)$.

We remark that the same construction can be iterated, so for any
non--negative integer $k$ we can recursively have a formula for
the rank polynomial of the lattice of all order ideals of a fence
with $k$ higher asymmetric peaks.

\section{Open problems and Conjectures}

In~\cite{MZ} using recursive formulas stated in Proposition~\ref{P:5}
it is proved that sequences $f_{n,k}$ and $c_{n,k}$ are indeed unimodal;
for definitions and comprehensive surveys about unimodal and (strong)
log--concave sequences we refer to~\cite{Bre2,Hog,Pro,Sta1,Wag}
(and the references therein).

We feel that the following stronger statement is true.

\begin{conj} \label{C:C}
For any $3 \ne n \in \mathbb{N} \setminus \{0\}$ and all $0 \leq k \leq n$,
the sequence $f_{n,k}$ is log--concave in $k$, viz.
$f_{n,k}^{2} \geq f_{n,k-1} f_{n,k+1}$ for any $k \in \[n-1\]$.

Moreover, for any $4 \leq n \in \mathbb{N}$ and all $0 \leq k \leq 2n$, the
sequence $c_{n,k}$ is strong log--concave in $k$, viz.
$c_{n,k}^{2} > c_{n,k-1} c_{n,k+1}$ for any $k \in \[2n-1\]$.
\end{conj}

Using a computer, Conjecture~\ref{C:C} has been verified for
distributive lattices of order ideals of fences and crowns, for
all fences and crowns with cardinality less or equal than $90$.

Moreover, we note that it would be of very great interest to study the
following much more general problem.

\begin{opb}
Characterize finite posets for which the distributive lattice of
order ideals is rank (strong) log--concave or just rank unimodal.
\end{opb}

\section*{Aknowledgment}

The author would like to thank Francesco Brenti for suggesting this problem
and his helpful advice, and Christian Krattenthaler for useful remarks.

redpajama_set_name: RedPajamaArXiv


Newly built contemporary villas situated in a select community of 14 homes in La Alqueria, Benahavis. There is a choice of three main designs and three sub-designs to complement personalised kitchens, bathrooms and overall decorative fittings, with additional space available to create a Spa, home cinema, entertainment area or gym. Also, private gardens and spacious terraced areas with different pool decks. This is an area dotted with luxurious villas and from its private setting enjoys an almost 360-degree panorama of golf courses, mountain scenery and the Mediterranean coastline from Marbella towards Estepona. On clear days the outlines of Gibraltar and the North African coastline can be seen, and as there are very many clear days here it is the perfect spot from which to explore the beautiful natural scenery that surrounds the area. Set roughly midway between the sea and the pretty white washed village of Benahavís - at under three kilometres from both - the gated community is peaceful yet on the doorstep of everything.


text:
Q: How to update attributes in an XML file with InstallScript Question:
I am struggling trying to figure out how to update some attributes in an XML file using InstallScript. I would like to use proper XML parsing functions, but I can't find anything that suggests this version of InstallScript is capable of it. How can I do this? Should I just attempt falling back on a combination of FileInsertLine and FileGrep? Is there a library I'm missing?
Background:
What software versions?
I am using InstallShield 11 on Windows Server 2003 R2.
Why am I not just using the existing 'XML File Changes' feature?
Because I am doing an upgrade and running into this bug. It affects the XML File Change feature because it is tied to a Component (well, that's my guess). I spent too long trying the official workaround, but couldn't coax it to work. I eventually found out it was much easier and more reliable to copy the files over using InstallScript + a single line batch file. It feels a bit hacky, but it is something that actually works.
Now I am trying to figure out the easiest and simplest-to-figure-out-years-later way to replicate the effects of the 'XML File Changes' feature in InstallScript.
Please let me know if you need any more information, I will be glad to provide it.
EDIT:
I ended up going with the InstallScript way to do it after all - it has tended to be the way that everything else in this installer project was implemented, and it looked (and turned out to be) pretty quick to do. I started out with the code shown by TheTraveler in that thread and modified it to suit my needs.
Here is the code:
prototype UpdateWebConfigAttributes();
prototype ReplaceValueOf(OBJECT, STRING, STRING);

function UpdateWebConfigAttributes()
    OBJECT oDoc, oNode;
    NUMBER i;
    STRING sWebConfigFilePath;
    BOOL successfulLoad;
begin

    sWebConfigFilePath = "Path\\To\\Web.config";

    if Is(FILE_EXISTS, sWebConfigFilePath) = FALSE then
        MessageBox("Could not find Web.config file.", 0);
    endif;

    // get values from public properties
    set oDoc = CreateObject("Msxml2.DOMDocument.4.0");
    if !IsObject(oDoc) then
        MessageBox("Could not create XML Document", 0);
        return -1;
    endif;

    oDoc.async = FALSE;
    oDoc.setProperty("SelectionLanguage", "XPath");

    successfulLoad = oDoc.load(sWebConfigFilePath);
    if !successfulLoad then
        MessageBox("Could not load Web.config as an xml file", SEVERE);
        return -1;
    endif;

    ReplaceValueOf(oDoc, "//add[@key=\"ConnectionDriver\"]", CONNECT_DRIVER);
    ReplaceValueOf(oDoc, "//add[@key=\"ConnectionType\"]", CONNECT_TYPE);
    ReplaceValueOf(oDoc, "//add[@key=\"ConnectionString\"]", CONNECT_STRING_WEBCONFIG);
    ReplaceValueOf(oDoc, "//add[@key=\"ShowConnection\"]", "False");

    oDoc.save(sWebConfigFilePath);
    set oDoc = NOTHING;
end;

function ReplaceValueOf(oDoc, xPath, valueToPutIn)
    OBJECT oNode;
begin
    set oNode = oDoc.selectNodes(xPath)(0);
    try
        oNode.attributes.getNamedItem("value").value = valueToPutIn;
    catch
        MessageBox("Could not set '" + xPath + "' with '" + valueToPutIn + "'", SEVERE);
    endcatch;
end;

A: I recall the XML Changes being pretty buggy back in those days.  What project type are you using?  If it's an MSI type then I would create a merge module in WiX and use it's XML changes pattern.  You can then add that merge module to your InstallShield project.
If you really do want to do it in InstallScript then you are going to have to use CoCreateObject() to call into an XML DOM.
There is a discussino and example over at XML Installscript
But again, I wouldn't do it this way.  I'd either upgrade to a newer version of InstallShield or I'd leverage the Util extension in WiX to keep it declarative.  I don't like writing these types of custom actions as it usually doesn't end well.
Util Schema


cgal-discuss - Example use of incident_cells?
Is there an example on using OutputIterator CGAL::Triangulation_3< Traits, TDS, SLDS >::incident_cells(Vertex_handle v,OutputIterator cells )const to write all cells incident to vertex v?
Re: Example use of incident_cells?
> to write all cells incident to vertex v?
You must provide an OuputIterator, so you can't directly pass "inc_cells"


text:
Harpers Ferry in Jefferson County, West Virginia — The American South (Appalachia)
Arsenal Square
By Bill Coughlin, November 10, 2006
1. Arsenal Square Marker
Arsenal Square. . Two brick arsenal buildings, which once housed about 100,000 weapons produced at the Harpers Ferry Armory, occupied these grounds. . . Capture of the firearms was the objective of John Brown's 1859 raid. . . Eighteen months after Brown's attack, the Civil War erupted. When Virginia militia advanced on the town on April 18, 1861, an outnumbered Federal garrison burned the arsenal and evacuated the town. The buildings and nearly 15,000 guns were destroyed. . This historical marker was erected by National Park Service. It is in Harpers Ferry in Jefferson County West Virginia
Two brick arsenal buildings, which once housed about 100,000 weapons produced at the Harpers Ferry Armory, occupied these grounds.
Capture of the firearms was the objective of John Brown's 1859 raid.
Eighteen months after Brown's attack, the Civil War erupted. When Virginia militia advanced on the town on April 18, 1861, an outnumbered Federal garrison burned the arsenal and evacuated the town. The buildings and nearly 15,000 guns were destroyed.
Erected by National Park Service.
Location. 39° 19.372′ N, 77° 43.803′ W. Marker is in Harpers Ferry, West Virginia, in Jefferson County. Marker is at the intersection of Shenandoah Street and Potomac Street, on the right when traveling north on Shenandoah Street. Marker is in the Lower Town at Harpers Ferry National Historical Park. Touch for map. Marker is in this post office area: Harpers Ferry WV 25425, United States of America. Touch for directions.
Other nearby markers. At least 8 other markers are within walking distance of this marker. "for the deposit of arms" (here, next to this marker); Burned, Flooded, and Leveled (a few steps from this marker); High Street in 1886
2. John Brown's Fort
The marker can be seen to the left of the fort in the Lower Town at Harpers Ferry National Historical Park.
(a few steps from this marker); Large Arsenal Foundation (within shouting distance of this marker); Large Arsenal (within shouting distance of this marker); John Brown (within shouting distance of this marker); "The War That Ended Slavery" (within shouting distance of this marker); Harper's Ferry History (within shouting distance of this marker). Touch for a list and map of all markers in Harpers Ferry.
More about this marker. The top of the marker contains a photograph of "The small arsenal in 1860. It is the building on the left."
Also see . . . Harpers Ferry National Historical Park. National Park Service. (Submitted on October 25, 2008, by Bill Coughlin of North Arlington, New Jersey.)
Categories. • Notable Places • War, US Civil •
By Craig Swain, May 17, 2009
The marker stands on the foundation trace for one of the arsenal buildings.
More. Search the internet for Arsenal Square.
Credits. This page was last revised on June 16, 2016. This page originally submitted on October 25, 2008, by Bill Coughlin of North Arlington, New Jersey. This page has been viewed 1,294 times since then and 5 times this year. Photos: 1, 2. submitted on October 25, 2008, by Bill Coughlin of North Arlington, New Jersey. 3. submitted on May 25, 2009, by Craig Swain of Leesburg, Virginia.


text:
Q: Absolutely continuous functions are dense in $L^1$ I would like to prove that $ac[0,1]$ which is defined as the set of all absolutely continuous functions on $[0, 1]$ is dense in $L^1([0,1])$.
I was thinking about defining, for $f \in L^1,$
$$f_n(x):=n \int_{(x,x+1/n)} |f|\ d\lambda$$
and prove that:
$1)\ f_n$ is absolutely continuous.
$2)\ f_n$ converge to $f$ in $L^1$ norm ($\|f_n-f\|_1<\epsilon$)
I know how to prove $1$, but I need help in $2.$ Thanks.

A: For the sake of convenience, extend $f$ to $\mathbb{R}$ by letting $f(x) = 0$ if $x \notin [0,1]$.  Given $n \in \mathbb{N}$, define $f_{n}(x) = n \int_{x}^{x + n^{-1}} f(y) \, \lambda(dy)$, where $\lambda$ is the Lebesgue measure.  Note that we don't want the absolute value bars in the integrand: if $f = -1$, for example, then $n \int_{x}^{x + n^{-1}} |f(y)| \, \lambda(dy) = 1$, from which we would obtain $\|f - f_{n}\|_{L^{1}([0,1])} \simeq 2$.
I'll trust that you can prove $(f_{n})_{n \in \mathbb{N}}$ is a sequence of absolutely continuous functions.
Now I'll prove $f_{n} \overset{L^{1}}\to f$.  The key fact is the following: if $f \in L^{1}(\mathbb{R})$ and $\tau_{y}f$ denotes the $L^{1}$-function $(\tau_{y}f)(x) = f(x + y)$, then
$$\tau_{y}f \overset{L^{1}}\to f \quad \text{as} \, \, y \to 0.$$
If $f \in C_{c}(\mathbb{R})$, then this follows from uniform continuity of $f$.  The general case then follows by density of $C_{c}(\mathbb{R})$ and uniform boundedness of the operators $\{\tau_{y}\}_{y \in \mathbb{R}}$ acting on $L^{1}(\mathbb{R})$.  Indeed,
$$\|\tau_{y}f\|_{L^{1}(\mathbb{R})} = \int_{-\infty}^{\infty} |f(x + y)| \, \lambda(dx) = \int_{-\infty}^{\infty} |f(x)| \, \lambda(dx) = \|f\|_{L^{1}(\mathbb{R})}$$
so $\|\tau_{y}\| = 1$.  If $f \in L^{1}(\mathbb{R})$ and $(h_{n})_{n \in \mathbb{N}} \subseteq C_{c}(\mathbb{R})$ satisfies $h_{n} \overset{L^{1}}\to f$, then
\begin{align*}
\limsup_{y \to 0} \|\tau_{y}f - f\|_{L^{1}(\mathbb{R})} &\leq \limsup_{y \to 0} \left[\|\tau_{y} h_{n} - h_{n} \|_{L^{1}(\mathbb{R})} + \|h_{n} - f\|_{L^{1}(\mathbb{R})} + \|\tau_{y}h_{n} - \tau_{y} f\|_{L^{1}(\mathbb{R})}\right] \\
&\leq 2 \|h_{n} - f\|_{L^{1}(\mathbb{R})}.
\end{align*}
Taking $n \to \infty$, we find $\limsup_{y \to 0} \|\tau_{y}f - f\|_{L^{1}(\mathbb{R})} = 0$, proving the claim.
Now to see that $f_{n} \overset{L^{1}}\to f$, integrate and change variables:
\begin{align*}
\|f - f_{n}\|_{L^{1}([0,1])} &= \int_{0}^{1} |f(x) - f_{n}(x)| \, \lambda(dx) \\
&= \int_{0}^{1} \left| f(x) - n \int_{0}^{n^{-1}} f(x + y) \, \lambda(dy) \right| \, \lambda(dx) \\
&= \int_{0}^{1} n \int_{0}^{n^{-1}} \left|f(x) - f(x + y) \right| \, \lambda(dy) \, \lambda(dx) \\
&= n \int_{0}^{n^{-1}} \left[ \int_{0}^{1} \left|f(x) - f(x + y) \right| \, \lambda(dx) \right] \, \lambda(dy) \\
&= n \int_{0}^{n^{-1}} \|\tau_{y} f - f \|_{L^{1}(\mathbb{R})} \, \lambda(dy) \\
&\leq \sup\{ \|\tau_{y} f - f\|_{L^{1}(\mathbb{R})} \, \mid \, y \in [0,n^{-1}]\}
\end{align*}
Since $\lim_{y \to 0} \|\tau_{y}f - f\|_{L^{1}(\mathbb{R})} =0$, we conclude
$$\lim_{n \to \infty} \|f_{n} - f\|_{L^{1}(\mathbb{R})} = 0.$$


text:
Notice re OF CONTINUING VIOLATION OF PRELIMINARY INJUNCTION DATED JUNE 21, 2007 by Xcentric Ventures, LLC(an Arizona Corporation), Ed Magedson(an individual) (Attachments: # 1 Exhibit A)(Speth, Maria) Modified on 11/13/2007 (TLJ). DOCUMENT DOES NOT CONTAIN A SIGNATURE, ATTORNEY NOTICED TO REFILE.
Xcentric Ventures, LLC et al v. Stanley et al Doc. 96 1 2 3 4 5 6 7 8 9 10 11 12 13 14 JABURG & WILK, P.C. ATTORNEYS AT LAW 3200 NORTH CENTRAL AVENUE SUITE 2000 PHOENIX, ARIZONA 85012 Kraig J. Marton, #003816 kjm@jaburgwilk.com Maria Crimi Speth, #012574 mcs@jaburgwilk.com Adam S. Kunz, #018827 ask@jaburgwilk.com Laura A. Rogal, #025159 lar@jaburgwilk.com JABURG & WILK, P.C. 3200 North Central Avenue, Suite 2000 Phoenix, Arizona 85012 (602) 248-1000 Attorneys for Plaintiffs IN THE UNITED STATES DISTRICT COURT DISTRICT OF ARIZONA XCENTRIC VENTURES, LLC, an Arizona corporation, d/b/a "RIPOFFREPORT.COM"; ED MAGEDSON, an individual Plaintiffs, v. WILLIAM "BILL" STANLEY, an individual; WILLIAM "BILL" STANLEY d/b/a DEFAMATION ACTION.COM; WILLIAM "BILL" STANLEY d/b/a COMPLAINTREMOVER.COM; WILLIAM "BILL" STANLEY aka JIM RICKSON; WILLIAM "BILL" STANLEY aka MATT JOHNSON; ROBERT RUSSO, an individual; ROBERT RUSSO d/b/a COMPLAINTREMOVER.COM; ROBERT RUSSO d/b/a DEFENDMYNAME.COM; ROBERT RUSSO d/b/a QED MEDIA GROUP, L.L.C.; QED MEDIA GROUP, L.L.C.; QED MEDIA GROUP, L.L.C. d/b/a DEFENDMYNAME.COM; QED MEDIA GROUP, L.L.C. d/b/a COMPLAINTREMOVER.COM; DEFAMATION ACTION LEAGUE, an unincorporated association; and INTERNET DEFAMATION LEAGUE, an unincorporated association; Defendants. 10297-14/LAR/LAR/620426_v1 Case No: 07-954-PHX-NVW NOTICE OF CONTINUING VIOLATION OF PRELIMINARY INJUNCTION DATED JUNE 21, 2007 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Case 2:07-cv-00954-NVW Document 96 Filed 11/09/2007 Page 1 of 3 Dockets.Justia.com 1 2 3 4 5 6 7 8 9 10 11 12 13 14 JABURG & WILK, P.C. ATTORNEYS AT LAW 3200 NORTH CENTRAL AVENUE SUITE 2000 PHOENIX, ARIZONA 85012 Plaintiffs Xcentric Ventures, LLC and Ed Magedson (collectively, "Plaintiffs"), hereby serve Notice of the following actions of Defendants William Stanley and Robert Russo in continuing violation of the Preliminary Injunction of this Court dated June 21, 2007: A short time after the hearing before this Court on Plaintiffs' Motion for Contempt, Defendant Robert Russo harassed a participant in Xcentric's Corporate Advocacy Program. Defendant Russo threatened to ruin the company's reputation and overwhelm the company with legal expenses because of the company's participation in the Corporate Advocacy Program. Rick Gelber is the President of the Taylor Wingate Group, a California-based corporation that participates in Xcentric's Corporate Advocacy Program. During the hearing before this Court on November 1, 2007, counsel for Plaintiffs referenced communications between Rick Gelber and Defendant Russo for the purpose of impeaching statements made by Defendant Russo. As a result of this line of testimony, it was disclosed that the Taylor Wingate Group participates in Xcentric's Corporate Advocacy Program. Negative postings about the Taylor Wingate Group that had been taken down prior to the hearing have since re-appeared on websites owned by Defendant William Stanley. The threats made by Defendant Russo against the Taylor Wingate Group are more specifically described in the Declaration of Rick Gelber, attached hereto as Exhibit "A". DATED this 9th day of November, 2007. JABURG & WILK, P.C. 15 16 17 18 19 20 21 22 23 24 25 26 27 28 2 10297-14/LAR/LAR/620426_v1 /Laura Rogal/ Kraig J. Marton Maria Crimi Speth Adam S. Kunz Laura Rogal Attorneys for Plaintiffs Page 2 of 3 Case 2:07-cv-00954-NVW Document 96 Filed 11/09/2007 1 2 3 4 5 6 7 8 9 10 11 12 13 14 JABURG & WILK, P.C. ATTORNEYS AT LAW 3200 NORTH CENTRAL AVENUE SUITE 2000 PHOENIX, ARIZONA 85012 Certificate of Service I hereby certify that on November 9, 2007, I electronically transmitted the attached document to the Clerk's Office using the CM/ECF System for filing, and for transmittal of a Notice of Electronic Filing to the following CM/ECF registrants: Teresa Kay Anderson Snell & Wilmer LLP One Arizona Center 400 E Van Buren Phoenix, AZ 85004 Michael Kent Dana Snell & Wilmer LLP 400 E Van Buren Phoenix, AZ 85004-0001 With a COPY of the foregoing emailed on the 9th day of November, 2007, to: William "Bill" Stanley videowebsites@yahoo.com With a COPY of the foregoing hand-delivered on the 9th day of November, 2007, to: Honorable Neil V Wake United States District Court District of Arizona /s/ Debbie Gower 15 16 17 18 19 20 21 22 23 24 25 26 27 28 3 10297-14/LAR/LAR/620426_v1 Case 2:07-cv-00954-NVW Document 96 Filed 11/09/2007 Page 3 of 3
Attachment 1


Comfort in Time of DistressPs 77 Hebrew title: A psalm by Asaph.
I cry aloud, and he hears me.
but I cannot find comfort.
when I meditate, I feel discouraged.
I am so worried that I cannot speak.
and remember years of long ago.
6I spend the night in deep thought;77.6 Some ancient translations deep thought; Hebrew song.
7"Will the Lord always reject us?
Will he never again be pleased with us?
8Has he stopped loving us?
Does his promise no longer stand?
9Has God forgotten to be merciful?
that God is no longer powerful."77.10 Verse 10 in Hebrew is unclear.
I will recall the wonders you did in the past.
I will meditate on all your mighty acts.
13Everything you do, O God, is holy.
No god is as great as you.
you showed your might among the nations.
the descendants of Jacob and of Joseph.
and the depths of the sea trembled.
and lightning flashed in all directions.
but your footprints could not be seen.
with Moses and Aaron in charge.


text:
Hot nourishing Meals on Wheels are especially welcome in winter
January 4th, 2016 7:10 AM
Some of the people who provide the Dunmanway Resource Centre meals on wheels service - from left - Saoirse O'Connell, TUS placement; Amy O'Brien, acting Older People Development Worker, and Valerie Hurley of the voluntary board of directors.
By Áilín Quinlan
TWICE a week for the past eight years, a fleet of cars leaves the Dunmanway Family Resource Centre, each vehicle packed with trays full of piping-hot dinners.
The two-course meals – a main course and dessert – are distributed by the centre's Meals on Wheels volunteers to about 60 people living in the town and its hinterland.
Aged 60 and over, these men and women enjoy the tasty home-cooked menu served up by the centre's excellent cooks.
Bacon and cabbage is the norm for Mondays says acting older people's services co-ordinator Amy O'Brien, who reports that this is hugely popular, while on Wednesdays it's usually a casserole, stews or shepherd's pie. The food is prepared by a roster of cooks who start work around 9.30am or earlier, in order to have the meals ready for 12.15pm when the fleet of ten volunteer drivers arrive to collect the food and deliver it all over the area.
Clients pay €4 for a dinner: 'Everything is prepared from scratch – people tell us it's a very good service and those who get it really value it. It can get a bit pressurised – it's a lot of food to prepare from scratch in the morning, so some of our cooks choose to come in a little earlier,' Amy reports.
The service which was pioneered by Dunmanway Geriatric Society more than 40 years ago, she explains, is a crucial one for people who cannot prepare food themselves: 'Some of our clients would be quite ill or very frail and may not be up to preparing a full dinner for themselves every day. It's been a great service for the community,' says Amy, who has just completed a Masters Degree in Community Development at NUI Galway.
As an extension of the Meals on Wheels Service, clients can also participate in a weekly Thursday Social Day, which includes morning tea and scones, a three-course lunch and a programme of activities, all for €10.
The Meals on Wheels service, which is supported to the tune of €28,000 by the HSE, actually costs in the region of €45,000.
The funding shortfall is made up through the meals income and the cost of the social day as well as a programme of fundraising activities, such as exercise classes, crafts or gardening, run by the Centre.
The Centre also runs a special men's lunch at noon on a Tuesday, costing €5 for dinner, dessert, tea and coffee
• Anybody who wishes to sign up to any of the Meals on Wheels services should contact their public health nurse or the Dunmanway Family Resource Centre at 023-8868110.
Skibbereen Community Alert Group relaunched
Covid-19 Thursday: 28 deaths, 3,955 new cases
Crosshaven to leave the lights on longer


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The filing in a federal court case also makes clear that Cheney was at the center of White House machinations rebutting criticism from former US Ambassador Joseph Wilson, who charged in summer 2003 that the Bush administration had "twisted" intelligence to justify invading Iraq in March 2003. While seeking to discredit Wilson, administration officials disclosed to reporters that Wilson's wife, Valerie Plame, worked for the CIA.
Just a week after the Defense Department announced plans to put the National Security Agency in charge of military cyber defense and attack, the agency's reach has already expanded to include monitoring of government civilian networks.
Given the NSA's involvement in the Bush administration's warrantless eavesdropping program, critics are concerned that the monitoring of government traffic on private-sector telecommunication networks that are used by the general public would allow the agency to once again spy on large swaths of non-government traffic without a warrant.
The Obama administration will proceed with a Bush-era plan to use National Security Agency assistance in screening government computer traffic on private-sector networks, with AT&T as the likely test site, according to three current and former government officials.
President Obama said in May that government efforts to protect computer systems from attack would not involve "monitoring private-sector networks or Internet traffic," and Department of Homeland Security officials say the new program will scrutinize only data going to or from government systems.
But the program has provoked debate within DHS, the officials said, because of uncertainty about whether private data can be shielded from unauthorized scrutiny, how much of a role NSA should play and whether the agency's involvement in warrantless wiretapping during George W. Bush's presidency would draw controversy.
TVNL Comment: Still waiting for that change we were promised.
A document filed in federal court this week by the Justice Department offers new evidence that former vice president Richard B. Cheney helped steer the Bush administration's public response to the disclosure of Valerie Plame Wilson's employment by the CIA and that he was at the center of many related administration deliberations.
The administration's discussion of Wilson's link to the CIA was meant to undermine criticism by her husband of administration allegations that Iraq attempted to acquire uranium, a matter that her husband had probed for the CIA, according to testimony presented in a 2007 trial.
Now two new laws in Colorado will allow many people to collect rainwater legally.
TVNL Comment: How is that for a free nation; your rain is owned by the government!


text:
Bush's New 'Axis of Evil'
September 2, 2021 February 11, 2011 by Patrick J. Buchanan
George W. Bush must have been the despair of the history department of every school his daddy managed to get him into.
Consider his latest excursion into the history of the republic, at Southern Methodist, where the Great Man's papers are to be housed.
"What's interesting about our country, if you study history, is that there are some 'isms' that occasionally pop up. One is isolationism and its evil twin protectionism and its evil triplet nativism. So if you study the '20s, for example, there was an American-first policy that said, 'Who cares what happens in Europe?' … And there was an immigration policy that I think during this period argued we had too many Jews and too many Italians, therefore we should have no immigrants. And my point is that we've been through this kind of period of isolationism, protectionism and nativism. I'm a little concerned that we may be going through the same period. I hope that these 'isms' pass."
First, "America First" was the antiwar movement begun in 1940 and backed by the young John F. Kennedy and his brother Joe, Gerald Ford and ex-president Herbert Hoover. It had nothing to do with the 1920s.
In the Harding-Coolidge decade, America was deeply interested in "what happens in Europe." It began with Hoover rushing U.S. food aid to the defeated nations of World War I and even to the USSR, for which Lenin personally thanked the Americans.
In 1921, President Harding called a Washington Naval Conference that produced the greatest disarmament treaty of modern times, in which America, Britain, France, Italy and Japan agreed to deep cuts and severe limits on the strategic weapons of the day, battleships.
In 1924, Charles G. Dawes advanced the Dawes Plan to ease the reparations burden on Germany, for which he won the Nobel Peace Prize.
In 1927, Coolidge convened a second naval disarmament conference to bring cruisers under the same limits as battleships — but the British balked.
In 1928 came the Kellogg-Briand Pact, by which scores of nations renounced war as an instrument of national policy. Undeniably utopian, it was hardly a mark of an isolationist America.
Secretary of State Frank Kellogg won our country's fourth Nobel Peace Prize.
In 1929 came the Young Plan to further ease a reparations burden on Germany then being exploited by the rising Nazi Party.
Wrote British historian A.J.P. Taylor: "American policy was never more active and never more effective in regard to Europe than in the 1920s. Reparations were settled; stable finances were restored; Europe was pacified, all mainly due to the United States."
What is Bush talking about, and why is he trashing this Republican record like some court historian of FDR?
As for "protectionism," Harding did approve the Fordney-McCumber Tariff Act of 1922, doubling rates to 38 percent. But he also slashed Woodrow Wilson's income tax rates by two-thirds, back to 25 percent.
Result: Unemployment, 12 percent when Harding took office, was 3 percent when Calvin Coolidge left. Manufacturing output rose 64 percent in the Roaring Twenties. Between 1923 and 1927, U.S. growth was 7 percent a year. At decade's end, America produced 42 percent of the world's goods.
Compare this economic triumph with the fruits of W's free-trade policy that wiped out 6 million U.S. manufacturing jobs, one of every three we had, and put America in hock to China.
The protectionism Bush calls "evil" was the policy of 12 Republican presidents from Abraham Lincoln to Coolidge, who made the GOP America's Party and converted this country into the industrial marvel of mankind.
Is Bush oblivious to this? Did someone at Phillips Academy, Yale or Harvard Business School tell him Lincoln, William McKinley and Theodore Roosevelt were free-traders?
As for "nativism," the term dates to the mid-19th century and had to do with hostility to Catholics and Irish, not Italians and Jews.
The 1924 Immigration Act, to end the Great Wave of the previous 30 years from Southern and Eastern Europe, did seek to preserve the ethnic character of the country. Yet, after 40 years of that moratorium, the Melting Pot having done its work, America was more united and socially at peace in the Eisenhower-JFK era than she has been before or since.
Is every immigration restriction law "evil," Mr. Bush? Are Japan and South Korea evil because they have never accepted immigration? Has mass immigration benefited Europe, where David Cameron and Angela Merkel are bewailing the disaster of "multiculturalism"? Is your successor, Gov. Rick Perry, evil for calling for troops on the border to stop the invasion you failed to halt?
For eight years, Bush pursued interventionism, free trade and open borders. Result: two wars that have bled his country and reaped a harvest of hate, the deindustrialization of America and a republic on its way to becoming the new world order's Tower of Babel.
Political result: A wipeout of the GOP in 2006 and 2008, and Bush going home to Texas with the lowest job approval in presidential history.
Bush ought to sue Phillips Academy for educational malpractice.
Categories Columns Tags America First, American History, Economic Nationalism, Free Trade, George Bush, Immigration, New World Order, War Post navigation
Ideology vs. the National Interest
Will Multiculturalism End Europe?


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