text:
Q: Foreign Key not showing in dropdown box | MVC I'm somewhat new to MVC and have been following along with a tutorial but it has no answers regarding my question. For my Create page, the Foreign keys are not showing up. Basically, on the Projects page I created a project, on the People page I created a person. So when I try to create a ProjectRole on the ProjectRoles page, the ProjectId and PersonId are not showing up in the drop-down menu. Down below all of my code, I have provided a screenshot of what I have tried to put into words.
My models:
public class Project
{
    public int Id { get; set; }

    [Required]
    [MaxLength(30)]
    public string Name { get; set; }

    [Required]
    public DateTime StartDate { get; set; }

    [Required]
    public DateTime DueDate { get; set; }

    public ICollection<ProjectRole> ProjectRoles { get; set; }
}

public class Person
{
    public int Id { get; set; }

    [Required]
    [MaxLength(30)]
    public string FirstName { get; set; }

    [Required]
    [MaxLength(30)]
    public string MiddleName { get; set; }

    [Required]
    [MaxLength(30)]
    public string LastName { get; set; }

    [Required]
    public string Email { get; set; }

    public ICollection<ProjectRole> ProjectRoles { get; set; }
}
public class ProjectRole
{
    public int Id { get; set; }

    [Required]
    public double HourlyRate { get; set; }

    [ForeignKey("Person")]
    public int PersonId { get; set; }

    [ForeignKey("Project")]
    public int ProjectId { get; set; }

    [ForeignKey("AppRole")]
    public int RoleId { get; set; }

}

My Controller code:
public IActionResult Create()
{
    return View();
}

[HttpPost]
[ValidateAntiForgeryToken]
public async Task<IActionResult> Create([Bind("Id,HourlyRate,PersonId,ProjectId,RoleId")] ProjectRole projectRole)
{
    if (ModelState.IsValid)
    {
        _context.Add(projectRole);
        await _context.SaveChangesAsync();
        return RedirectToAction(nameof(Index));
    }
    return View(projectRole);
}

And my view code here:
@model Project2.Models.Entities.ProjectRole

@{
    ViewData["Title"] = "Create";
}

Create

ProjectRole

        <form asp-action="Create">

                <input asp-for="HourlyRate" class="form-control" />

                <select asp-for="PersonId" class ="form-control" asp-items="ViewBag.PersonId"></select>

                <select asp-for="ProjectId" class ="form-control" asp-items="ViewBag.ProjectId"></select>

                <input asp-for="RoleId" class="form-control" />

                <input type="submit" value="Create" class="btn btn-primary" />

        </form>

    Back to List

@section Scripts {
    @{await Html.RenderPartialAsync("_ValidationScriptsPartial");}
}

Screenshot of example of what I mean:

A: I suppose more personable to display the Person Name (and the Project Name) in the select controls, but to pass the PersonId (and ProjectId) to the Create method when click on the Create button.
Therefore, prepare the Person list (and the Project list) like below:
public IActionResult Create()
{
    var persons = new List<SelectListItem>();
    // Iterate through `Persons`
    foreach(var p in _context.Persons)
    {
        persons.Add(new SelectListItem() { Value= p.Id, Text = p.FirstName+", "+p.LastName});
    }
    ViewBag.Persons = persons;

    // Prepare the projects list (like `Persons` list above)
    // ... your code here
    ViewBag.Projects = persons;

    return View(new ProjectRole(){ /* Add your code here to create the ProjectRole .../* });
}

And in the view:

    <select asp-for="PersonId" class="form-control" asp-items="ViewBag.Persons"></select>

    <select asp-for="ProjectId" class="form-control" asp-items="ViewBag.Projects"></select>

For additional information see The Select Tag Helper
*NOTE: And I would recommend to create compound view model to include all required information. *

*

*ViewModels or ViewBag?

*Understanding Best Way to Use Multiple Models in ASP.NET MVC


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Tourism and Entertainment Minister, Hon. Dr. Wykeham McNeill (2nd right), cuts the ribbon to formally open the newly refurbished Berrydale rafting station and loading bay in Portland on Wednesday (April 2). The loading bay, which was refurbished at a cost of $2.8 million, is one of four projects undertaken in the parish by the Tourism and Entertainment Ministry at a cost of over $48 million, with funding provided by the Tourism Enhancement Fund (TEF). Assisting the Minister (from left) are: Port Antonio Mayor, Councillor Benny White; Portland's Custos Rotulorum, Hon. Lincoln Thaxter; Member of Parliament for East Portland, where the facility is located, Dr. Lynvale Bloomfield; Tourism and Entertainment State Minister, Hon. Damion Crawford.
Portland's residents and visitors should benefit significantly from four major projects undertaken at a cost of over $48 million, to boost tourism in the parish.
The projects include repairs to over three kilometres of roadway between the communities of Fellowship and Berrydale.
Funding was provided by the Tourism Enhancement Fund (TEF), and project management undertaken by the Tourism Product Development Company (TPDCo).
The projects, carried out by the Ministry of Tourism and Entertainment, include repairs to over three kilometres of roadway between the communities of Fellowship and Berrydale, at a cost of $20 million; refurbishing of the Neville Antonio Park in Port Antonio – $13 million; completion of the Portland Craft Village development, also in Port Antonio – $12.5 million; and repairs to the Berrydale rafting jetty and loading bay – $2.8 million.
There were also inputs from the Portland Parish Council and National Works Agency (NWA), which carried out work on the craft village and Berrydale road, respectively.
The developments are consistent with the Government's strategic job and economic growth priority, focusing on development and growth in key productive sectors (tourism and rural development); integration of culture and entertainment into the economy; and improved roadways.
Tourism and Entertainment Minister, Hon. Dr. Wykeham McNeill, and State Minister, Hon. Damion Crawford, accompanied by parish officials and stakeholders, toured the projects on April 2. He then symbolically handed them over to the TPDCo and Council, which will manage them, during a ceremony at Neville Antonio Park.
In his remarks, Dr. McNeill commended the participating stakeholders for their inputs in implementing "such well needed infrastructural improvements," while highlighting their importance.
The Minister said he is confident that the work carried out on the projects will redound to the benefit of the many Jamaicans who live in, or visit Portland. "These four projects…and the many others undertaken across the country, truly exemplify how the tourism sector benefits the average Jamaican," he added.
Meanwhile, Executive Director of the TEF, Clyde Harrison, said the entity welcomes the end result of the money spent in Portland.
Also lauding the projects were Portland's Custos, Hon. Lincoln Thaxter; Member of Parliament for East Portland, where the developments are located, Dr. Lynvale Bloomfield; Port Antonio's Mayor, Councillor Benny White; and Portland Craft Traders Association President, Joan Shaw.


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Instead of soaking in the sweetness of her newborn son, Marlee Weir was praying he would suffer only hearing loss.
That's because sometime before birth, her son Reece became infected with cytomegalovirus (CMV). After a normal pregnancy, he was born severely jaundiced and covered in little purple spots. Marlee and her husband Casey were shaken, but the Brandon residents quickly learned there could be bigger problems ahead.
CMV is a common virus that causes mostly cold symptoms in healthy adults, but can cause a host of problems in babies infected in utero. According to the Centers for Disease Control, about 1 in 5 infants with a congenital CMV infection like Reece's will suffer more permanent disabilities including developmental disabilities or hearing loss.
"I prayed a lot that day and the following days as we camped out in the neonatal intensive care unit that if something had to be wrong, for it to be hearing because I knew the technology available and knew that we could address hearing loss quickly," she said.
Over the next two weeks, she began to think her prayers had been answered.
"I kind of had mother's intuition that he was not hearing because he wouldn't startle to loud noises," Marlee said.
A few weeks after his birth, Reece had a hearing screening at the University of Mississippi Medical Center and her suspicion was confirmed – he had severe to profound hearing loss in both ears.
A 2010 graduate of UMMC's School of Health Related Professions, Marlee Weir's training as an occupational therapist kicked in. She knew learning during the first three years of a child's life was crucial and immediately began researching Reece's options.
"I know that the younger you start things with kids, the better the outcomes will be," she said.
One of the biggest health issues facing Mississippi and the nation is obesity, yet the way the problem is measured remains debated.
While the Centers for Disease Control and Prevention has popularized the Body Mass Index, a formula using a person's height and weight as an effective way to identify obesity, the location of body fat isn't taken into account.
"The fat that's in your abdomen -- the visceral fat around the organs -- is what's important in increasing risk for cardiovascular disease and diabetes," said Dr. John Hall, chair of the Department of Physiology and Biophysics and lead investigator for the University of Mississippi Medical Center's Mississippi Center for Obesity Research.
Most of the chronic diseases for which obesity is a risk factor are related to the visceral, or deep, fat, he said. The subcutaneous fat, the kind found under the skin in areas such as the thighs or buttocks, may not cause as many problems.
A case of bad timing a few years ago stopped Dr. Douglas Bacon from interviewing for the top position in the Department of Anesthesiology at the University of Mississippi Medical Center.


I want to use decibels to describe compression ratio. So for example ratio 4:1 means that every 4 input decibels above threshold causes raise of output only 1 decibel. And I also want ratio to be constant on whole range above threshold. So for 4:1, if input is 8 decibels above I want output raise 2 decibels.
How to make my ratio which is described in decibels, to make it constant?
I suppose I need to modify that equation somehow, but have no idea how?
Browse other questions tagged audio dynamic-range-compression or ask your own question.


text:
Genevieve Nnaji Stars In Hollywood Film
Nigeria's screen diva, Genevieve Nnaji is starring in a movie, "Farming" alongside Hollywood actors Kate Beckinsale, Damon Idris, and Gugu Mmbatha-Raw.
The film which is being shot in the U.K. and Nigeria started production in August and is being directed by Nigerian-British writer and director, Adewale Akinnuoye-Agbaje.
The film is based on Adewale's autobiographical story, "Farming." Genevieve, who shared few pictures of herself on set on her Instagram is popular for classic movies including, "Ije" and "Keeping Faith".
The Nollywood veteran recently made her directorial debut with her upcoming movie, "Lion Heart". The script was developed at the Sundance Film Institute and chronicles his own coming-of-age story growing up fostered by a white working-class family in the U.K.
In the film, the lead character is sent to the U.K. by his parents in hopes of achieving a better life. Instead, he becomes the feared leader of a white skinhead gang in 1980s England. When all seems lost and he spirals into self-destruction, a benevolent teacher (played by Mbatha-Raw) offers him one last chance at redemption.
Genevieve won the Africa Movie Academy Award for Best Actress in a Leading Role in 2005, making her the first actress to win the award.
She has starred in over 80 movies and is considered to be one of the best paid actresses in Nollywood.
Due to her contributions to the Nigeria movie industry, she became the first actress to be awarded Best Actress at the 2001 City Peoples Awards, the award ceremony that previously had only recognized politicians and business conglomerates.
She is also the first actress to be awarded as Best Actress by the Census Board of Nigeria in 2003.
In 2009, she was referred to as the Julia Roberts of Africa by Oprah Winfrey.
In 2004, Nnaji signed a recording contract with EKB Records, a Ghanaian record label, and released her debut album `One Logologo line', a mix of R&B and Urban music.
In November 2015, Nnaji produced her first movie called "Road to Yesterday'' which won Best Movie Overall-West Africa at the 2016 Africa Magic Viewers Choice Awards.
SOURCE: VANGUARD

redpajama_set_name: RedPajamaCommonCrawl


There was a dearth of women in the wilderness of Michigan, and Chauncey and John traveled back to Cato, New York for wives. Between them, the two men married five women, becoming brothers-in-law three times in the process. The resultant 24 children, some of whom didn't survive to adulthood, had confusing familial relationships where siblings were also cousins and a man could be both the father and an uncle to his children.


text:
\section{Introduction}
Let us denote $[n]:=\{1,\ldots, n\}$ and $2^{[n]}:=\{X: X\subset [n]\}$. A {\it family} $\mathcal{F}\subset 2^{[n]}$ of subsets of $[n]$ is called {\it intersecting}, if any two of its sets intersect. A family is called {\it $k$-uniform}, if it consists of sets of size $k$. For shorthand, we denote such a family by $\mathcal{F}\subset {[n]\choose k}$.

The investigation of intersecting families started from the following famous result due to Erd\H{o}s, Ko and Rado.

\begin{theorem}[{\cite{ErdHos1961}}]\label{thm:ekr}
  Let $n \geq 2k$ and consider an intersecting family $\mathcal{F}\subset {[n]\choose k}$. Then
  \begin{align*}
    |\mathcal{F}|\le \binom{n-1}{k-1}.
  \end{align*}
  For $n > 2k$ equality occurs if and only if $\mathcal{F}$ consists of all $k$-sets
  that contain a fixed element of $[n]$.
\end{theorem}
An intersecting family with all its sets containing a fixed element of $[n]$ is called {\it trivial}.
The EKR theorem was sharpened by Hilton and Milner \cite{HM}, who determined the size of the largest {\it non-trivial} intersecting $k$-uniform family. Later, a very strong result in this direction was obtained by Frankl \cite{Fra1}. Below we give a stronger and a more convenient version of that theorem, proven in \cite{KZ}.  The \textit{degree $\delta(x)$} of $x \in [n]$ is defined as the number of members
of $\mathcal{F}$ that contain $x$. The {\it maximal degree} of $\mathcal{F}$ is denoted by $\Delta(\mathcal{F})$. The {\it diversity} $\gamma(\mathcal{F})$ of $\mathcal{F}$ is the number of sets from $\mathcal{F}$ not containing an element of the maximal degree: $\gamma(\mathcal{F}):=|\mathcal{F}|-\Delta(\mathcal{F})$.

\begin{theorem}[\cite{KZ}]\label{thmKZ} Let $n>2k>0$ and $\mathcal{F}\subset {[n]\choose k}$ be an intersecting family. Then, if $\gamma(\mathcal{F})\ge {n-u-1\choose n-k-1}$ for some real $3\le u\le k$, then \begin{equation}\label{eq01}|\mathcal{F}|\le {n-1\choose k-1}+{n-u-1\choose n-k-1}-{n-u-1\choose k-1}.\end{equation}
\end{theorem}

Other {\it stability} results for $k$-uniform intersecting families were obtained by several researchers (cf. \cite{BBN, DF, EKL, KM, KK, Pyad, R}). Dinur and Friedgut \cite{DF} introduced the methods of analysis of Boolean functions to the study of intersecting families. Roughly speaking, they showed that {\it any} intersecting family is essentially contained in {\it juntas} with small centers. We say that a family $\mathcal J\subset 2^{[n]}$ is a {\it $j$-junta}, if there exist a subset $J\subset [n]$ of size $j$, such that the membership of a set in $\mathcal{F}$ is determined only by its intersection with $J$, that is, for some family $\mathcal J^*\subset 2^{J}$ we have $\mathcal{F}=\{F:F\cap J\in \mathcal J^*\}$. Here is one of their two main results.

\begin{theorem}[\cite{DF}]\label{thmdf} For any integer $r\ge 2$, there exist functions $j(r), c(r)$, such that for any integers $1 < j(r) < k < n/2$, if $\mathcal{F}\subset {[n]\choose k}$ is an intersecting family with $|\mathcal{F}|\ge c(r){n-r\choose k-r}$, then there exists an intersecting $j$-junta $\mathcal J$ with
$j\le j(r)$ and \begin{equation}\label{eqDF} |\mathcal{F}\setminus\mathcal J|\le c(r){n-r\choose k-r}.\end{equation}
\end{theorem}
The methods of Dinur and Friedgut were developed and extended to other extremal questions on set systems by other researchers, notably Ellis, Keller, and Lifshitz \cite{EKL, KL}.\\

In this paper we study intersecting families with respect to another natural measure of non-triviality: the distribution of degrees. The question we address in this paper is as follows: how large an intersecting family may be, provided that all elements of the ground set have the same degree? We call such families {\it regular}. In what follows, we always denote the degree of an element in such a family  by $\delta$.\\

Theorem \ref{thm:ekr} implies that for $n>2k$
the largest $k$-uniform intersecting family has one element of degree $\binom{n-1}{k-1}$
while all the other elements of $[n]$ have degree $\binom{n-2}{k-2}$. Hence, the spread between the largest and the smallest degree is very large.

Using Theorem~\ref{thmKZ} with $u=3$, we get that any family of size strictly bigger than ${n-1\choose k-1}+{n-4\choose k-3}-{n-4\choose k-1}=3{n-3\choose k-2}+{n-3\choose k-3}$ has diversity strictly smaller than ${n-4\choose k-3}$. Therefore, the maximal degree of any such family is $\Delta(\mathcal{F})=|\mathcal{F}|-\gamma(\mathcal{F})> 3{n-3\choose k-2}$. At the same time, the sum of the degrees of all elements except for a most popular one is $(k-1)\Delta(\mathcal{F})+k\gamma(\mathcal{F})$, and so the minimal degree of $\mathcal{F}$ is at most $\frac {k-1}{n-1}\Delta(\mathcal{F})+\frac {k}{n-1}\gamma(\mathcal{F})<\frac {k-1}{n-1}\Delta(\mathcal{F})+\frac{k}{n-1}{n-4\choose k-3}<\frac {k-1}{n-1}\Delta(\mathcal{F})+{n-3\choose k-2}$. It is easy to see that the difference between the maximal and the minimal degrees is at least $\Delta (\mathcal{F})-\frac {k-1}{n-1}\Delta(\mathcal{F})-{n-3\choose k-2}=\frac{n-k}{n-1}\Delta(\mathcal{F})-{n-3\choose k-2}>\frac 3{2}{n-3\choose k-2}-{n-3\choose k-2}>0$. Therefore, we may conclude the following.

\begin{proposition}\label{prop1} If $n\ge 2k>0$, then any regular $k$-uniform intersecting family has size at most $3{n-3\choose k-2}+{n-3\choose k-3}$.
\end{proposition}

We write $S_n$ for the symmetric group on $[n]$. A \textit{symmetric family} $\mathcal{F}$ on $[n]$ is a family such that the automorphism group $Aut(\mathcal{F}):=\{\sigma\in S_n: \sigma(\mathcal{F})=\mathcal{F}\}$ of $\mathcal{F}$ is a transitive subgroup of $S_n$, that is, for all $i,j\in [n]$ there exists a permutation $\sigma\in Aut(\mathcal{F})$ such that $\sigma(i)=j$. Clearly, any symmetric family must be regular, and the converse is not true in general.

Cameron, Frankl and Kantor \cite{Cameron1989} studied maximal intersecting families $\mathcal{F}\subset 2^{[n]}$ (that is, families of size $2^{n-1}$), that are additionally symmetric or regular.
They proved the following result that relates the size of the ground set and the size of the smallest set in the family.

\begin{theorem}[{\cite[Theorem 3]{Cameron1989}}]\label{thm:bnd_on_n} Consider an intersecting family $\mathcal{F}\subset 2^{[n]}$ of size $2^{n-1}$ and an arbitrary set $F\in \mathcal{F}$.
  \begin{enumerate}
   \item If $\mathcal{F}$ is symmetric, then $n \le |F|^2$.
   \item If $\mathcal{F}$ is regular, then $n \le \frac{2}{\pi}4^{|F|}$.
  \end{enumerate}
\end{theorem}

Ellis, Kalai and Narayanan investigated $k$-uniform symmetric intersecting families
\cite{Ellis2017}. They obtained the following analogue of the theorem above.

\begin{theorem}[{\cite[Lemma 4.5]{Ellis2017}}]\label{thm:ellis_proj}
  Let $k^2-k+1$ be prime.
  A $k$-uniform symmetric intersecting family $\mathcal{F}$ on $[n]$ satisfies $n \leq k^2-k+1$.
  Further, equality holds if and only if $\mathcal{F}$ is a point-transitive projective plane of order $k-1$.
\end{theorem}

This result was already shown by Lov\'{a}sz without the characterization of equality, but not published
(it is mentioned in \cite{Lovasz1975,Lovasz1977}), and F\"{u}redi \cite{Furedi1981,Furedi1990}
for regular families. We give a second proof for this result that was obtained independently.

\begin{theorem}[{\cite[Corollary 3]{Furedi1981}}]\label{thm:char_n_max}
  A $k$-uniform regular intersecting family $\mathcal{F}$ on $[n]$ satisfies $n \leq k^2-k+1$.
  Further, equality holds if and only if $\mathcal{F}$ is a projective plane of order $k-1$.
\end{theorem}

Ellis, Kalai and Narayanan obtained the following bound on the size of a $k$-uniform symmetric intersecting family.

\begin{theorem}[{\cite[Theorem 1.3]{Ellis2017}}] \label{thm:ellis_bnd}
  There exists a constant $c > 0$ (independent of $n$ and $k$) such that for $k \leq n/2$ and any symmetric intersecting family $\mathcal{F}$ we have
  \begin{align*}
     |\mathcal{F}| \leq \exp\left( -\frac{c(n-2k) \log n}{k(\log n - \log k)} \right) \binom{n}{k}.
  \end{align*}
\end{theorem}
Their proof uses tools from the analysis of Boolean functions, notably, the sharp threshold result due to Friedgut and Kalai \cite{FrKa}.
They managed to show that this bound is
tight up to the constant $c$ in the exponent, if $k/n$ is bounded away from zero.

Using Theorem~\ref{thmdf}, we are able to obtain the following upper bound on the size of regular intersecting families.

\begin{theorem}\label{thmmain1} For any $r\in \mathcal N$ there exists a constant $C=C(r)$, such that for $n>Ck$ any regular family $\mathcal{F}\subset {[n]\choose k}$ satisfies
$$|\mathcal{F}|\le C {n-r\choose k-r}.$$
\end{theorem}

Moreover, it is possible to strengthen the theorem above to the following setting. We say that the family is {\it $\alpha$-irregular} for $\alpha\ge 1$, if the ratio between the maximal degree and the minimal degree is at most $\alpha$.

\begin{theorem}\label{thmmain2} For any $r\in \mathcal N$ and $\alpha\in \mathbb R_{\ge 1}$ there exists a constant $C=C(r,\alpha)$, such that for $n>Ck$ any $\alpha$-irregular family $\mathcal{F}\subset {[n]\choose k}$ satisfies
$$|\mathcal{F}|\le C {n-r\choose k-r}.$$
\end{theorem}

The next result uses algebraic methods, and may be applied for all values of $n$ and $k$. However, it works only for regular families and is weaker than Theorem~\ref{thmmain1} in the range when Theorem~\ref{thmmain1} may be applied.

\begin{theorem}\label{thm:bnd_n_small_intro}
  A regular $k$-uniform intersecting family on $[n]$ satisfies
  \begin{align*}
    |\mathcal{F}| \leq \frac{\binom{n}{k}}{1 + \frac{(n-k)(n-k-1)(n-k-2)}{k(k-1)(k-2)}}.
  \end{align*}
\end{theorem}

We only know sporadic examples for which the bound is tight. We also remark that it is not difficult to see that the bound in Theorem~\ref{thm:bnd_n_small_intro} is always stronger than the bound from Proposition~\ref{prop1}.

If $n=ck$ for some constant $c$, then all the upper bounds we give have order $\Theta({n-1\choose k-1})$. This is in sharp contrast with the situation for symmetric intersecting families, for which in \cite{Ellis2017} the upper bounds have order $o({n-1\choose k-1})$ for $n-2k=\Theta(k/\log k)$. However, this is not a shortcoming of our methods: for any $c$ we give a series of examples of a regular intersecting families in ${[n]\choose k}$, such that $n>ck$ and that have size $\Theta({n-1\choose k-1})$.
Therefore, the following is true.

\begin{proposition}\label{prop3} If $k=o(n)$, then any regular intersecting family in ${[n]\choose k}$ has size $o({n-1\choose k-1})$. At the same time, for any $c>0$ there are regular intersecting families in ${[n]\choose k}$ with $n>ck$ and which have size $\Theta_c({n-1\choose k-1})$.
\end{proposition}

We note that a similar construction was used in \cite[Theorem 3]{Furedi1981A} by F\"uredi and \cite[Theorem 2.4]{Frankl1986} by Frankl and F\"uredi. However, their studies were concerned with intersecting families with bounded maximum degree and they did not require the families to be regular. In particular, the lower bounds they prove resemble Theorem~\ref{thmmain1}, but are only meaningful for $n\gg k^2$, which is by Theorem~\ref{thm:char_n_max} not interesting as long as regular intersecting families are considered.

The structure of the paper is as follows. In Section~\ref{sec:prelim} we review some notions from the theory of association schemes, needed in the proofs of Theorems~\ref{thm:char_n_max} and \ref{thm:bnd_n_small_intro}. In Section~\ref{sec3} we prove the upper bounds on the size of regular intersecting families and some of its generalizations, in particular, to the case of degrees of subsets. In the first part of Section~\ref{sec4} we provide some general constructions of regular intersecting families, which in particular imply the corresponding part of Proposition~\ref{prop3}. In the second part of Section~\ref{sec4} we study in detail the case $n=2k$, and give some results in the case $n=2k+1$. Finally, in Section~\ref{sec5} we conclude and give future directions for research.

\section{The Johnson Scheme}\label{sec:prelim}

In the proof of Theorems~\ref{thm:char_n_max} and \ref{thm:bnd_n_small_intro} we use the theory of association schemes.
To make this paper more self-contained, this section summarizes the necessary notions and results.
We refer to Delsarte's PhD thesis \cite{Delsarte1973} as a standard reference on
the combinatorial applications of association schemes.

\begin{definition}
 Let $X$ be a finite set. A $k$-class association scheme is a pair $(X, \mathcal{R})$,
 where $\mathcal{R} = \{ R_0, \ldots, R_k \}$ is a set of symmetric binary relations
 on $X$ with the following properties:
 \begin{enumerate}[(a)]
  \item $\{ R_0, \ldots R_k \}$ is a partition of $X \times X$.
  \item $R_0$ is the identity relation.
  \item There are constants $p_{ij}^\ell$ such that for $x, y \in X$ with $(x, y) \in R_\ell$
  there are exactly $p_{ij}^\ell$ elements $z$ with $(x, z) \in R_i$ and $(z, y) \in R_j$.
 \end{enumerate}
\end{definition}
We denote $p^0_{ii}$ by $r_i$, the \text{valency of $R_i$}. We denote $|X|$ by
$v = \sum_{i=0}^k r_i$. The relations $R_i$ can be described by their adjacency matrices
$A_i \in \mathbb{C}^{v \times v}$ defined by
\begin{align*}
  (A_i)_{xy} = \begin{cases}
                1 & \text{ if } x R_i y,\\
                0 & \text{ otherwise.}
               \end{cases}
\end{align*}
Let $\mathbf{J}$ denote the all-ones matrix and let $\mathbf{j}$ denote the all-ones vector.
It is easily verified that the matrices $A_i$ are Hermitian and commute pairwise,
hence we can diagonalize them simultaneously, i.e. there is a set of common eigenvectors.
From this we obtain pairwise orthogonal, idempotent Hermitian
matrices $E_j \in \mathbb{C}^{v \times v}$ with the properties (possibly after reordering)
\begin{align}
  &\sum_{j=0}^k E_j = \mathbf{I}, && E_0 = v^{-1} \mathbf{J},\notag\\
  &A_i = \sum_{j=0}^k P_{ji} E_j, && E_j = v^{-1} \sum_{i=0}^k Q_{ij} A_i \label{eq:decomp_adj}
\end{align}
for some constants $P_{ij}$ and $Q_{ij}$.
From this it is clear that the
matrices $A_0, A_1, \ldots, A_k$ have $k+1$ common eigenspaces $V_0 = \langle \mathbf{j} \rangle,
V_1, \ldots, V_k$, where the eigenspaces $V_j$ has dimension $f_j := \text{rk}(E_j)=\text{tr}(E_j)$
and $P_{ji}$ is the eigenvalue of  $A_i$ in the eigenspace $V_j$.
Note that $E_j$ is an orthogonal projection from $\mathbb{C}^{v \times v}$
onto $V_j$. There are several connections between the parameters of an association
scheme, for example $r_i = P_{0i}$. For us the following is important.

\begin{lemma}[{\cite[Eq. (4.34)]{Delsarte1973}}]\label{lem:rel_P_Q}
  We have $r_i Q_{ij} = f_j P_{ji}$.
\end{lemma}

Let $\mathcal{F}$ be a subset of $X$.
Let $\chi$ be the \textit{characteristic vector} of $\mathcal{F}$, i.e.
\begin{align*}
  \chi_x = \begin{cases}
	      1 & \text{ if } x \in \mathcal{F},\\
	      0 & \text{ otherwise. }
           \end{cases}
\end{align*}
Define the \textit{inner distribution} $a$ of $\mathcal{F}$, $|\mathcal{F}| > 0$, by
\begin{align*}
  a_i = \frac{1}{|\mathcal{F}|} \chi^T A_i \chi.
\end{align*}

The essential property of association schemes, which we use, is summarized in
the following result which is often referred to as \textit{Delsarte's linear
programming bound (LP bound)}.

\begin{theorem}[{\cite[Lemma 2.5.1 (iv) and Prop. 2.5.2]{Brouwer1989}}]\label{thm:lp_bound}
  We have $(aQ)_j = \frac{v}{|\mathcal{F}|}\chi^T E_j \chi \geq 0$ with equality if
  and only if $\chi \in V_j^\perp$.
\end{theorem}
The vector $aQ$ is often referred to as the \textit{MacWilliams transform} of $a$.
A very particular, well-known case in Delsarte's LP bound is when $\chi \in \langle \mathbf{j} \rangle + V_j$
for some $j \in \{ 1, \ldots, k \}$:

\begin{lemma}\label{lem:char_single_es}
Let $\chi$ be the characteristic vector of $\mathcal{F}$.
If $\chi \in \langle \mathbf{j} \rangle + V_j$, then the following holds:
\begin{enumerate}[(a)]
 \item If $F \in \mathcal{F}$, then $F$ meets exactly $|\mathcal{F}| (P_{0i} - P_{ji})/ v + P_{ji}$
elements of $\mathcal{F}$ in relation $R_{i}$.
 \item If $F \notin \mathcal{F}$, then $F$ meets exactly $|\mathcal{F}| (P_{0i} - P_{ji})/ v$
elements of $\mathcal{F}$ in relation $R_{i}$.
\end{enumerate}
\end{lemma}
\begin{proof}
  Let $\psi$ be the characteristic vector of $\{ F \}$. Write
  $\chi = \frac{|\mathcal{F}|}{v} \mathbf{j} + \chi'$ with $\chi' \in V_j$.
  Then the number of elements in $\mathcal{F}$ that are in relation $R_i$ to $F$ is
  \begin{align*}
    \psi^T A_i \chi &= \psi^T \left( \sum_{i=0}^k P_{ji} E_j \right) \Big( \frac{|\mathcal{F}|}{v} \mathbf{j} + \chi'\Big) \\
    &= \psi^T \Big(P_{0i} \frac{|\mathcal{F}|}{v} \mathbf{j} + P_{ji} (\chi - \frac{|\mathcal{F}|}{v} \mathbf{j})\Big) = |\mathcal{F}| (P_{0i} - P_{ji})/ v + P_{ji} \psi^T \chi. \qedhere
  \end{align*}
\end{proof}

The \textit{Johnson scheme} has $X = {[n]\choose k}$. Two $k$-sets are in relation $R_i$ if their intersection is of size $k-i$. We recall the eigenvalues of the Johnson scheme.
As this association scheme is \textit{cometric}, its eigenspaces have a canonical ordering which we use in the following.

\begin{lemma}[{\cite[p. 48]{Delsarte1973} and \cite[Theorem 6.5.2]{Godsil2015}}]\label{lem:evs}
  The eigenvalue of $A_i$ of the eigenspace $V_j$ is
  \begin{align*}
   P_{ji}
    &= \sum_{h=0}^i (-1)^{h} \binom{j}{h} \binom{k-j}{i-h} \binom{n-k-j}{i-h}\\
    &= \sum_{h=i}^k (-1)^{h-i+j} \binom{h}{i} \binom{n-2h}{k-h} \binom{n-h-j}{h-j}.
  \end{align*}
  In particular, $P_{jk} = (-1)^{j} \binom{n-k-j}{k-j}$.
\end{lemma}

We will use Theorem \ref{thm:lp_bound} for $j=1$ and $j=2$, so we need $Q_{i1}$
and $Q_{i2}$ explicitly.

\begin{lemma}\label{lem:dual_evs}
 We have
 \begin{enumerate}[(a)]
  \item $c_1 Q_{i1} = kn-in-k^2$, and
  \item $c_2 Q_{i2} = (k-i)(k-i-1)(n-k-i)(n-k-i-1) - 2i^2(k-i)(n-k-i) + i^2(i-1)^2$,
 \end{enumerate}
 where $c_1 = k(n-k)/f_1$ and $c_2 = k(k-1)(n-k)(n-k-1)/f_2$.
\end{lemma}
\begin{proof}
  Evaluate Lemma \ref{lem:rel_P_Q} for the stated cases using Lemma \ref{lem:evs}.
\end{proof}

The eigenspaces of the Johnson scheme have various nice combinatorial descriptions.
Let $\mathcal{G}_S$ denote the family of all $k$-sets
that contain a fixed $s$-set $S$. Clearly, $|\mathcal{G}_S| = \binom{n-s}{k-s}$. Let $\lambda_s = \binom{n-s}{k-s}/\binom{n}{k}$.
Let $\psi_S$ denote the characteristic vector of $\mathcal{G}_S$.

\begin{lemma}[{\cite[Theorem 6.3.3]{Godsil2015}}]\label{lem:char_first_es}
  Fix $s \in \{ 1, \ldots, k \}$. The set $\cup_{r=1}^s\{ \psi_R - \lambda_r \mathbf{j}: R \in \binom{[n]}{r} \}$ spans $V_1+V_2+\ldots+V_s$.
\end{lemma}

We say that a family $\mathcal{F}$ is {\it $s$-subset-regular} if every $s$-set lies in the same number of elements $\delta_s$ of $\mathcal{F}$.
For any $r$ with $1 \leq r \leq s$, double counting pairs $( R, S ) \in \binom{[n]}{r} \times \binom{[n]}{s}$ with $R \subset S$
shows that every $r$-set lies in exactly $\delta_s \binom{n-r}{s-r}/\binom{k-r}{s-r}$ elements of $\mathcal{F}$.
The following is well-known, but we include a short proof for completeness.

\begin{lemma}\label{lem:char_F}
  Fix $s \geq 1$.
  Let $\mathcal{F}$ be a family of $k$-uniform sets and let $\chi$ be the characteristic vector of $\mathcal{F}$.
  Then (a) implies (b) and (c):
  \begin{enumerate}[(a)]
   \item $\mathcal{F}$ is $s$-subset-regular.
   \item The vector $\chi$ is orthogonal to the eigenspaces $V_1, \ldots, V_s$.
   \item The inner distribution $a$ of $\mathcal{F}$ satisfies
    \begin{align*}
      0 = \sum_{i=0}^k Q_{i1} a_i.
    \end{align*}
  \end{enumerate}
\end{lemma}
\begin{proof}
  The eigenspaces
  $V_0$, $V_1$, \ldots, $V_k$ are pairwise orthogonal,
  so we can write $\chi = E_0 \chi + (E_1+E_2+\ldots+E_k) \chi = \frac{|\mathcal{F}|}{v} \mathbf{j} + \chi'$,
  where $\chi' \in V_1 + V_2 + \ldots + V_k$.
  Hence, for any $R \in \binom{[n]}{r}$, where $1 \leq r \leq s$,
  \begin{align*}
    \chi^T \psi_R &= \left(\frac{|\mathcal{F}|}{v} \mathbf{j} \right)^T \left(\lambda_r \mathbf{j}\right)
		      + \chi'^T \left(\psi_R - \lambda_r \mathbf{j} \right)\\
		  &= |\mathcal{F}| \lambda_r + \chi'^T \left(\psi_R - \lambda_r \mathbf{j}\right).
  \end{align*}
  Since, by simple double counting, the degree of $R$ in $\mathcal{F}$ is $\lambda_r |\mathcal{F}|$, we have $\chi'^T \left(\psi_R - \lambda_r \mathbf{j} \right) = 0$.
  Hence, $\chi$ is orthogonal to all vectors in $V_1 + V_2 + \ldots + V_s$. Hence, (a) implies (b).

  By Theorem \ref{thm:lp_bound}, $\chi \in V_1^\perp$ is equivalent to (c).
\end{proof}

\section{Upper Bounds}\label{sec3}

\subsection{Proof of Theorems~\ref{thmmain1} and \ref{thmmain2}}

Clearly, Theorem~\ref{thmmain1} is implied by Theorem~\ref{thmmain2}, so we only need to prove Theorem~\ref{thmmain2}.  In terms of Theorem~\ref{thmdf}, we put $C=\max\{2\alpha j(r), 2c(r)\}$.  Fix $n,k$, such that $n\ge Ck$ and an $\alpha$-irregular family $\mathcal{F}\subset {[n]\choose k}$. Then Theorem~\ref{thmdf} states that there exists an intersecting $j$-junta $\mathcal J=\{A\in{[n]\choose k}: A\cap J\in \mathcal J^*\}$ for some $J$ with $|J|=j\le j(r)$ and $\mathcal J^*\subset 2^J$, such that $|\mathcal{F}\setminus \mathcal J|\le c(r){n-r\choose k-r}$.

We may assume that $|\mathcal{F}|> C{n-r\choose k-r}\ge 2 c(r){n-r\choose k-r}$, otherwise we have nothing to prove. Therefore, we have $|\mathcal{F}\cap \mathcal J|/|\mathcal{F}|> 1/2$.

Given the approximation by the junta $\mathcal J$, let us bound the degrees of elements in $\mathcal{F}$. On the one hand, clearly, any set from $\mathcal{F}\cap \mathcal J$ intersects $J$, therefore, the maximal degree of an element in $J$ is at least $|\mathcal{F}\cap \mathcal J|/j$. On the other hand, the average degree of an element in $[n]$ is $\frac kn|\mathcal{F}|\le|\mathcal{F}|/C.$ Hence, the ratio between the maximal and the minimal degree in $\mathcal{F}$ is at least $$\frac{|\mathcal{F}\cap \mathcal J|/j}{|\mathcal{F}|/C}>\frac {C}{2j}\ge \alpha. $$
We conclude that, if $|\mathcal{F}|>C{n-r\choose k-r}$, then $\mathcal{F}$ is not $\alpha$-irregular. Theorem~\ref{thmmain2} is proved. \\

\subsection{Proof of Theorem \ref{thm:bnd_n_small_intro}}

Theorem \ref{thm:bnd_n_small_intro} is implied by the following
result for $s$-subset-regular $k$-uniform intersecting families. It is a variation of Hoffman's bound.

\begin{theorem}\label{thm:bnd_n_small}
  Fix odd $s\ge 1$. An  $s$-subset-regular $k$-uniform intersecting family $\mathcal{F}$ on $[n]$ satisfies
  \begin{align*}
    |\mathcal{F}| \leq \frac{v}{1 - P_{0k}/P_{(s+2)k}} = \frac{\binom{n}{k}}{1 + \frac{\binom{n-k}{k}}{\binom{n-k-s-2}{k-s-2}}}.
  \end{align*}
  For $n > 2k$ we have equality only if the characteristic vector $\chi$ of $\mathcal{F}$
  lies in the span of $\mathbf{j}$ and the eigenspace $V_{s+2}$.
\end{theorem}
\begin{proof}[Proof of Theorem \ref{thm:bnd_n_small}]
  Let $\chi$ be the characteristic vector of $\mathcal{F}$.
  First notice that
  \begin{align}
    |\mathcal{F}| = \chi^T \chi = \frac{|\mathcal{F}|^2}{v} + \sum_{j=1}^k \chi^T E_j \chi.\label{eq:Y_Ysq}
  \end{align}
  Let $A = A_k - \sum_{i=1}^s P_{ik} E_i$.
  Obviously, $A$ has the eigenvalues $P_{jk}$ for $j = 0$ and $j > s$, and the eigenvalue $0$ for the eigenspace $V_i$, where $1 \leq i \leq s$.
  From the definition of $\mathcal{F}$, $\chi^T A_k \chi = 0$.
  By Lemma \ref{lem:char_F}, $\chi^T E_i \chi = 0$ for $1 \leq 1 \leq s$.
  Hence, $\chi^T A \chi = 0$.
  Notice that, using Lemma \ref{lem:evs} and Equation \eqref{eq:decomp_adj}, $P_{(s+2)k}$ is the smallest eigenvalue of $A$, so we obtain
  \begin{align*}
    0 = \chi^T A \chi
    &= \frac{P_{0k}}{v} \chi^T \mathbf{J} \chi + \sum_{j=s+1}^k P_{jk} \chi^T E_j \chi\\
    &\geq \frac{P_{0k}}{v} \chi^T \mathbf{J} \chi + P_{(s+2)k} \sum_{j=1}^k \chi^T E_j \chi\\
    &\stackrel{\eqref{eq:Y_Ysq}}{=} \frac{P_{0k}}{v} \cdot |\mathcal{F}|^2 + P_{(s+2)k} \left( |\mathcal{F}| - \frac{|\mathcal{F}|^2}{v} \right).
  \end{align*}
  Rearranging yields
  \begin{align*}
    |\mathcal{F}| \leq \frac{v}{1 - P_{0k}/P_{(s+2)k}},
  \end{align*}
  which, together with Lemma \ref{lem:evs}, shows the first half of the assertion.
  For the second half of the assertion notice that we have equality in this bound
  only if $\chi^T E_{s+2} \chi = \sum_{j=1}^k \chi^T E_j \chi$.
\end{proof}

Evaluating Lemma \ref{lem:char_single_es} for the case of equality in Theorem \ref{thm:bnd_n_small} and $s=1$, we
obtain that a $k$-set $F \notin \mathcal{F}$ meets exactly
$$\frac{3k(k-1)(k-2)}{n^2-3kn-n+3k^2}$$
elements of $\mathcal{F}$ in $k-1$ elements. As this has to be an integer, Theorem \ref{thm:bnd_n_small}
cannot be tight for many combinations of $n$ and $k$. In particular, the bound is never
tight for $3k(k-1)(k-2) < n^2-3kn-n+3k^2$.
For odd $s \geq 1$ fixed, the same argument yields that the bound is not tight for $n$ at least $\sim \sqrt{s+2} k^{\frac{s+2}{s+1}}$.

Ray-Chaudhuri and Wilson showed in \cite[Theorem 1]{RCW1975} that a $2t$-design contains at least $\binom{n}{t}$ blocks.
Together with Theorem \ref{thm:bnd_n_small} this implies that $s$-subset-regular intersecting families do not exist
when $s$ is large compared to $k$ (vaguely $k < \frac{3}{2} (s+1)$).

\begin{remark}
  The bound of
  \begin{align*}
    \frac{v}{1 - P_{0k}/P_{(s+2)k}}
  \end{align*}
  and its consequences hold for various other association schemes, where the eigenspaces can be
  characterized in a similar way. One example for such an association scheme would be the
  $q$-analog of the Johnson scheme, the $q$-Johnson (Grassmann) scheme.
\end{remark}

\subsection{Proof of Theorem \ref{thm:char_n_max}}

For small $n$ Delsarte's linear programming (LP) bound
corresponds to Theorem \ref{thm:bnd_n_small}, but for $n$ close to $k^2-k+1$ Delsarte's LP bound is much better.
In particular, Delsarte's LP bound implies Theorem \ref{thm:char_n_max}.
In the following we prove this formally.

\begin{proof}[Proof of Theorem \ref{thm:char_n_max}]
  The inner distribution $a$ of an intersecting family $\mathcal{F}$ has the form
  $a = (1, a_1, \ldots, a_{k-1}, 0)$.
  In terms of Lemma~\ref{lem:dual_evs}, denote $\alpha_i:=c_2 Q_{i2}$
  and $\beta_i:=c_1Q_{i1}$.
  By Lemma \ref{lem:char_F}, we know that $\sum_{i=0}^{k-1} \beta_i a_i = 0$.
  By Theorem \ref{thm:lp_bound}, we know that $\sum_{i=0}^{k-1} \alpha_i a_i \geq 0$.
  Hence, $\sum_{i=0}^{k-1} (\alpha_i - (k-1)(n-k-1)\beta_i)a_i \geq 0$.
  Using Lemma~\ref{lem:dual_evs}, we have that
  \begin{align*}
    \gamma_i := \alpha_i - (k-1)(n-k-1)\beta_i = -i(n-2)(kn-in-k^2+i).
  \end{align*}
  If $n > k^2-k+1$, then $\gamma_i < 0$ for all $i \in \{ 1, \ldots, k-1\}$.
  Hence, $\sum_{i=0}^{k-1} \gamma_i a_i \geq 0$ implies $a = (1, 0, \ldots, 0)$, which is a contradiction.
  Thus, we can assume $n = k^2-k+1$. Then $\gamma_i = 0$ for $i \in \{ 0, k-1 \}$
  and $\gamma_i < 0$ for $i \in \{ 1, \ldots, k-2 \}$.
  Hence, $\sum_{i=0}^{k-1} \gamma_i a_i \geq 0$ implies $a = (1, 0, \ldots, 0, a_{k-1}, 0)$.
  Now $\sum_{i=0}^{k-1} \beta_i a_i = 0$ together with Lemma~\ref{lem:char_F} implies $(k-1)k = a_{k-1}$.
  Therefore, $|\mathcal{F}| = k^2-k+1$, each element in $[k^2-k+1]$ has degree $k$, and each set in $\mathcal{F}$ meets all other sets in $\mathcal{F}$ in exactly one point.
  This is a well-known definition of a projective plane of order $k-1$.
\end{proof}

A similar method also gives us the following lower bound on the size of a
regular $k$-uniform intersecting family $\mathcal{F}$.

\begin{lemma}\label{lem:lower_bnd}
  If a regular $k$-uniform intersecting $\mathcal{F}$ exists, then
  \begin{align*}
   |\mathcal{F}| \geq 1 + \frac{k(n-k)}{k^2-n}
  \end{align*}
  with equality only if $\sum_{i=1}^{k-2} a_i = 0$.
\end{lemma}
\begin{proof}
  Let $a$ be the inner distribution of $\mathcal{F}$.
  By Lemma \ref{lem:char_first_es}, $kn-k^2 + \sum_{i=1}^{k-1} (kn-in-k^2)a_i =0$.
  It is easy to see that under this constraint $|\mathcal{F}| = \sum_{i=0}^{k-1} a_i$
  is minimized if $a = (1, 0, \ldots, 0, a_{k-1}, 0)$.
\end{proof}

\section{Lower Bounds}\label{sec4}

We have already shown that there are no regular families for $n>k^2-k+1$ and that for $n=k^2-k+1$  a projective plane of order $k-1$ is the only possible example. We note  that the existence of projective planes is only known for $k-1$ being a prime power. In the first part of this section we give some general constructions of how to build bigger regular intersecting families out of smaller ones, and derive some general lower bounds. In the second part of this section we discuss the case $n=2k$ and $n=2k+1$, as well as some results for small values of $k$.

We note that Ellis, Kalai, and Lifshitz in \cite[Section 4]{EKL} construct symmetric (and thus regular) $k$-uniform intersecting families on $[n]$ for all $k$ and $n$ satisfying $k\ge 1.1527\sqrt{n}$. They rely on the relation between intersecting families and difference covers for $\mathbb Z_n$. We could not improve their results for the case of regular intersecting families, so a very interesting question that remains is whether regular intersecting families exist for all sufficiently large $n,k$ satisfying $k\ge (1+o(1))\sqrt{n}$.

\subsection{General Constructions and Proposition~\ref{prop3}}
For any regular family $\mathcal{F}\subset {[n]\choose k}$ we put $\alpha(\mathcal{F}):=k/n$. We call this parameter the {\it ratio} of $\mathcal{F}$.
Due to the simple equality $|\mathcal{F}|k = n \delta$, $\alpha(\mathcal{F}) = \delta/|\mathcal{F}|$ and $\alpha(\mathcal{F})$ tells us, what proportion of all sets from $\mathcal{F}$ contains a given element.

Let us start with giving some general ways of constructing a regular intersecting family.
Assume that we have a regular intersecting family $\mathcal{F}\subset{X\choose k}$ on the set $X$, a regular intersecting family $\mathcal{F}_2\subset {Y\choose k_2}$ on the set $Y$, and a regular (but not necessarily intersecting) family $\mathcal{G}\subset {Z\choose m}$ on the set $Z$. Assume additionally that the sets $X,Y,Z$ are pairwise disjoint and that the ratios of $\mathcal{F}$ and $\mathcal{G}$ are the same: $\alpha(\mathcal{F}) = \alpha(\mathcal{G})$.

Fix a parameter $l$, $1\le l< \min\{|F_1\setminus F_2|:F_1,F_2\in \mathcal{F}, F_1\ne F_2\}$. Then the following families are intersecting and regular:
\begin{align}
  &\mathcal{F}^l:=\Big\{F'\in {X\choose k+l}: F'\supset F\text{ for some }F\in \mathcal{F}\Big\}, && \mathcal{F}^l\subset{X\choose k+l};\\
\label{eq11} &\mathcal{F}+\mathcal{G} :=\big\{F\cup G: F\in \mathcal{F}, G\in \mathcal{G}\big\}, &&\mathcal{F}+\mathcal{G}\subset {X\cup Z\choose k+m};\\
&\mathcal{F}\times \mathcal{F}_2:=\big\{F\times F_2:F\in \mathcal{F}, F_2\in \mathcal{F}_2\big\}, &&\mathcal{F}\times\mathcal{F}_2\subset {X\times Y\choose kk_2}.
\end{align}
It is easy to see that each of the families above is intersecting. Let us show that all these families are regular. It is obvious in the case of $\mathcal{F}\times \mathcal{F}_2$.

This is the most difficult to show in the case of family $\mathcal{F}^l$. First, note that, due to the choice of $l$, for any $F'\in \mathcal{F}^l$ there is a {\it unique} set $F\in \mathcal{F}$, such that $F'\supset F$. Thus, we may partition $\mathcal{F}^l$ into families $\mathcal{F}^l(F):=\{F'\in {X\choose k+l}: F'\supset F\}, $ where $F\in \mathcal{F}$. The elements in $\mathcal{F}^l(F)$ have two possible degrees. Each element $x\in F$ has degree $|\mathcal{F}^l(F)| = {|X|-k\choose l}$, while any element $x\in X\setminus F$ has degree ${|X|-k-1\choose l-1}$. It is clear that, since $\mathcal{F}$ is regular, each $x\in X$ has degree of the first type in $\alpha(\mathcal{F})|\mathcal{F}|$ families $\mathcal{F}^l(F)$ and the degree of the second type in all the others. Since the degree of each element in $\mathcal{F}^l$ is the sum of its degrees in $\mathcal{F}^l(F)$, we conclude that $\mathcal{F}'$ is regular. We note that $|\mathcal{F}^l| = {|X|-k\choose l}|\mathcal{F}|$.

The family $\mathcal{F}+\mathcal{G}$ is regular, since each element from $X$ is contained in $\alpha(\mathcal{F})$-fraction of all sets in $\mathcal{F}+\mathcal{G}$, and each element from $Y$ is contained in $\alpha(\mathcal{G})$-fraction of all sets in $\mathcal{F}+\mathcal{G}$. However, $\alpha(\mathcal{F}) = \alpha(\mathcal{G})$ by assumption. We note that $|\mathcal{F}+\mathcal{G}| = |\mathcal{F}||\mathcal{G}|$ and that $\alpha(\mathcal{F}+\mathcal{G}) = \alpha(\mathcal{F})$.\\

We use Construction \eqref{eq11} to show the lower bound from Proposition~\ref{prop3}. Fix $c>0$ from the proposition. Recall that we need to construct a regular intersecting family with ratio at most $1/c$ and size $\Theta({n-1\choose k-1})$. In particular, $\binom{n}{k}$ and $\binom{n-1}{k-1}$ are of the same order of magnitude. Find a prime power $q$, such that $(q^2+q+1)/(q+1)>c$. Put $\mathcal{F}$ to be the projective plane of order $q$ on the set $X$. Note that $\mathcal{F}$ is regular, intersecting, and $\alpha(\mathcal{F}) = (q+1)/(q^2+q+1)$. Take $\mathcal{G}={Z\choose l(q+1)}$, where $Z$ is a set of size $l(q^2+q+1)$ disjoint with $X$. Here we think of $q$ that is fixed and $l$ that tends to infinity. Then $\alpha(\mathcal{G})=\alpha(\mathcal{F})$. Thus, the family $\mathcal{F}+\mathcal{G}$ is regular, intersecting, and satisfies $\alpha(\mathcal{F}+\mathcal{G})<1/c$. Moreover, $\mathcal{F}+\mathcal{G}\subset  {|X\cup Z|\choose (l+1)(q+1)}$ and
$$\frac{|\mathcal{F}+\mathcal{G}|}{{|X\cup Z|\choose (l+1)(q+1)}} = \frac{(q^2+q+1){l(q^2+q+1)\choose l(q+1)}}{{(l+1)(q^2+q+1)\choose (l+1)(q+1)}}>\frac{(q^2+q+1){l(q^2+q+1)\choose l(q+1)}}{{l(q^2+q+1)\choose l(q+1)}\frac{\big((l+1)(q^2+q+1)\big)^{q^2+q+1}}{\big(lq^2\big)^{q^2} \big(l(q+1)\big)^{q+1}}}=\Theta_q(1).$$

\subsection{Case $n=2k$}

In the case $n=2k$ no intersecting family of $k$-sets can have size bigger than  $\binom{2k-1}{k}$.
Brace and Daykin \cite{Brace1972} (in somewhat different terms) showed that for $k$ not a power of $2$, this bound is tight.

\begin{theorem}[{\cite[Example 1]{Brace1972}}]\label{thm25}
  If $k$ is not a power of $2$, then a largest $k$-uniform regular intersecting
  family on $[2k]$ has size $\binom{2k-1}{k}$.
\end{theorem}

Brace and Daykin also noted that it is easy to see that if $k$ is a power of $2$
that then a regular family of size $\binom{2k-1}{k}$ would have a non-integer degree,
that is there the bound of $\binom{2k-1}{k}$ is not tight. We complete their result
by showing the following.

\begin{theorem}\label{thm:neq2k_k_power_of_2}
  If $k\ge 4$ is a power of $2$, then the largest $k$-uniform regular intersecting
  family on $[2k]$ has size $\binom{2k-1}{k}-3$.
\end{theorem}
Note that there is no regular intersecting family for $k=2$, $n=4$, and substituting $k=2$ in the bound above gives $0$. We also remark that from the proof we give below it is fairly easy to reconstruct the proof of Theorem~\ref{thm25}.
\begin{proof} First we show the upper bound. Let $\mathcal{F}$ be the largest regular intersecting family for $n=2k$, $k=2^t$. Consider the family $\mathcal{F}(2k):=\{F\in\mathcal{F}:2k\in F\}$. Simple counting shows that $|\mathcal{F}(2k)| = \frac{1}{2} |\mathcal{F}|$. Assume that the degree of an element $i$ in $\mathcal{F}(2k)$ is $d(i)$. If all $d(i)$, $i=1,\ldots, 2k-1$, are the same, then $\frac 12|\mathcal{F}|=|\mathcal{F}(2k)| = d(1)\frac{2k-1}{k-1}$, and thus $2(2k-1)$ must divide $|\mathcal{F}|$ and $|\mathcal{F}|\le {2k-1\choose k}-(2k-1)$. (Here we use an easy-to-check fact that $(2k-1)$ divides ${2k-1\choose k}$, but $2$ does not.)

Otherwise, consider the family \begin{equation}\label{eq38}\bar\mathcal{F}(2k):={[2k-1]\choose k}\setminus \big\{[2k] \setminus F:F\in \mathcal{F}(2k)\big\}.\end{equation}
Clearly, $\mathcal{F}\setminus \mathcal{F}(2k)\subset \bar\mathcal{F}(2k)$. Moreover, the degree of the element $i\in[2k-1]$ in $\bar \mathcal{F}(2k)$ is $\bar d(i) = {2k-2\choose k-1}-|\mathcal{F}(2k)|+d(i)$. Therefore, the degree of $i\in[2k-1]$ in $\mathcal{F}':=\mathcal{F}(2k)\cup \bar\mathcal{F}(2k)$ is \begin{equation}\label{eq39}{2k-2\choose k-1}-|\mathcal{F}(2k)|+2d(i).\end{equation}

We have $|\mathcal{F}| = {2k-1\choose k}-(2l+1)$ for some $l\ge 0$. Thus  $\mathcal{F}\setminus \mathcal{F}(2k)=\bar\mathcal{F}(2k)\setminus \mathcal G$, where $|\mathcal G|=2l+1$.  Since not all $d(i)$, $i=1,\ldots, 2k-1$, are equal to each other, by the displayed formula above two degrees in $\mathcal{F}'$ differ by at least $2$. Thus, one has to delete at least two sets from $\mathcal{F}'$ to get a regular family, which implies that $l\ge 1$ and proves $|\mathcal{F}|\le {2k-1\choose k}-3$.
This shows the upper bound in Theorem \ref{thm:neq2k_k_power_of_2}.\\

Now let us show the lower bound. The following lemma is the key step. (We postpone the proof of the lemma until the end of the proof of the theorem.)

\begin{lemma}\label{lem666} There exists a family $\mathcal Q\subset{[2k-1]\choose k-1}$ of size $\frac 12({2k-1\choose k-1}-3)$ with the degrees $d'(i)$ of $i\in [2k-1]$ satisfying $d'(1)-1 = d'(2)-1 = \ldots = d'(3k/2)-1 = d'(3k/2+1) = \ldots =  d'(2k-1)$ and such that the sets $A_1:=[k+1,2k-1]$, $A_2:=[1,k/2]\cup [3k/2+1,2k-1]$, and $A_3:=[k/2+1,k]\cup [3k/2+1,2k-1]$ are not in $\mathcal Q$.
\end{lemma}

Double counting the sum of $d'(i)$ in such $\mathcal Q$, we get that the degree of, say, $2k-1$ equals \begin{equation}\label{eq40}\frac{1}{2k-1}\Big(\frac {k-1}2\Big[{2k-1\choose k-1}-3\Big]-\frac {3k}2\Big)=\frac{k-1}{2(2k-1)}{2k-1\choose k-1}-\frac 32 = \frac 12{2k-2\choose k-2}-\frac 32,\end{equation}
which, in particular, is an integer number (recall that ${2k-1\choose k-1}$ is odd, and so is ${2k-2\choose k-2} = \frac {k-1}{2k-1}{2k-1\choose k-1}$). This shows that there are no divisibility obstructions to the existence of $\mathcal Q$.

 We put $\mathcal{F}(2k):=\{\{2k\}\cup Q: Q\in \mathcal Q\}$. Next, we consider the family $\bar\mathcal{F}(2k)$ defined as in \eqref{eq38}. The family $\mathcal{F}':=\mathcal{F}(2k)\cup\bar\mathcal{F}(2k)$ has size ${2k-1\choose k}$ and the degrees of the first $3k/2$ elements are bigger by two than the degrees of the last $k/2$ elements as we will see in the following. The degree of $2k$ in $\mathcal{F}'$ is $|\mathcal{F}(2k)| = \frac 12({2k-1\choose k}-3)$, which is the same as the degree of, say, $2k-1$, equal by \eqref{eq39} and \eqref{eq40} to
$${2k-2\choose k-1}-\frac 12\left({2k-1\choose k-1}-3\right)+{2k-2\choose k-2}-3 = \frac 12\left({2k-1\choose k-1}-3\right).$$
Moreover, $\mathcal{F}'$ contains $[2k-1]\setminus A_i$ for $i=1,2,3$. The family $\{[2k-1]\setminus A_i:i = 1, 2, 3\}$ has degree $2$ on the first $3k/2$ elements and $0$ on the remaining ones. Thus, deleting these sets from $\mathcal{F}'$, we get a regular family $\mathcal{F}$ of size ${2k-1\choose k}-3$.

It remains to show that $\mathcal{F}'$ is intersecting. Clearly, $\mathcal{F}(2k)$ is intersecting as all
elements contain $2k$ and $\bar\mathcal{F}(2k)$ is intersecting as all its elements are in $\binom{[2k-1]}{k}$.
By the definition of $\bar\mathcal{F}(2k)$ in \eqref{eq38}, if $F \in \mathcal{F}(2k)$, then $[2k] \setminus F \notin \bar\mathcal{F}(2k)$.
Hence, each element of $\mathcal{F}(2k)$ intersects all elements of $\bar\mathcal{F}(2k)$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem666}] Due to \eqref{eq40}, we only need to show that there exists $\mathcal Q$ of desired size and without $A_i$, in which any two degrees differ by at most $1$.

We start with a construction of some auxiliary family $\mathcal Q'$ which satisfies $|\mathcal Q'| = |\mathcal Q|$. The first part of $\mathcal Q'$ is the family $\mathcal Q(2k-1)$ of all sets containing the element $2k-1$ except for the sets $A_i$. This accounts for ${2k-2\choose k-2}-3$ sets, and we have to choose $\frac 12({2k-1\choose k-1}-3)-{2k-2\choose k-2}+3 = \frac 1{2k-2}{2k-2\choose k-2}+\frac 32$ sets more. The remaining sets we choose out of ${[2k-2]\choose k-1}$ in such a way that the degrees of
the elements in $[2k-2]$ differ by at most one and are non-decreasing as $i\in[2k-2]$ increases. This could be done as follows (the same idea was used in the proof of Theorem 1 in \cite{Brace1972}).

We start with any family $\mathcal Q''$ in ${[2k-2]\choose k-1}$ of needed size. Then, if there are $g,h\in [2k-1]$, such that the degree of $g$ is bigger than the degree of $h$ by at least $2$, then we take a set $A$, such that $g\in A$, $h\notin A$, and $A':=\{h\}\cup A\setminus \{g\}$ is not in the family, and replace $A$ with $A'$. We call such an operation a {\it $(g,h)$-replacement}. Repeating this procedure will eventually lead to a family $\mathcal Q(\overline{2k-1})$, in which any two degrees differ by at most $1$.
We may also w.l.o.g. assume that elements are ordered from the ones with the smaller degree to the ones with the larger degree.

The resulting family is $\mathcal Q':=\mathcal Q(\overline{2k-1})\cup \mathcal Q(2k-1)$. Due to the fact that $A_i\notin \mathcal Q(2k-1)$, the degrees of the first $3k/2$ elements are bigger by 2 than the degrees of $3k/2+1,\ldots, 2k-2$. However, the degree of any $i\in [3k/2]$ in $\mathcal Q'$ is at most
$${2k-3\choose k-3}-1+\Big\lceil\frac 12\big(\frac 1{2k-2}{2k-2\choose k-2}+\frac 32\big)\Big\rceil\le\frac{k-\frac32}{2k-2}{2k-2\choose k-2}+\frac 12.$$
where the first term in the left hand side is the degree of $1$ in $\mathcal Q(2k-1)$ and the second term is the upper integer part of the average degree in $\mathcal Q(\overline{2k-1})$. The right hand side is at most the right hand side of \eqref{eq40} plus $1$ (which should be the degree of $i\in[3k/2]$ in $\mathcal Q$), since the difference is $\frac 1{4k-4}{2k-2\choose k-2}-\frac 54\ge 0$ for any $k\ge 4$. Similarly, the degree of each element in $[3k/2+1,2k-2]$ in $\mathcal Q'$ is at most what it should be in $\mathcal Q$.

This implies that we can obtain the family $\mathcal Q$ with degrees that differ by at most $1$ doing $(2k-1,h)$-replacements (and thus transferring the excess of the degree of $2k-1$ to other elements). In particular, since $2k-1\in A_i$, it implies that such $\mathcal Q$ will not contain $A_i$, $i=1,2,3$. The lemma is proven.
\end{proof}

\subsection{The cases $n=2k+1$ and $k=4$}

Consider the case $n = 2k+1$. The bound in Theorem \ref{thm:bnd_n_small} in this case is $\frac{k-2}{2k-1} \binom{2k+1}{k}$.
For $k=3$ this is $7$ and Theorem \ref{thm:char_n_max} states that this is only obtained
if $\mathcal{F}$ is the Fano plane. It is easy to see that the bound is always an integer
and that the degree in case of equality is the integer $\frac{k-2}{2k-1} \cdot \frac{k}{2k+1} \binom{2k+1}{k} = \frac{k-2}{2k-1} \binom{2k}{k-1}$.
We found examples that reach this bound for $k \in \{3, 4\}$. For $k=3$ there is one example, the Fano plane.
By computer, we classified all examples for $k=4$ where there are exactly two regular examples of size $36$ up to isomorphism.
The following table summarizes our knowledge for small $k$ and $n = 2k+1$.

\begin{center}
\begin{tabular}{lrrrrr}
 $k$ & 3 & 4 & 5 & 6 \\ \midrule
 Theorem \ref{thm:bnd_n_small} & $7$ & $36$ & $154$ & $624$  \\ \midrule
 Largest & $7$ & $36$ & $\geq 110$ & $\geq 442$ \\ \midrule
 $\delta$ & $3$ & $16$ & $\geq 50$ & $\geq 204$ \\
\end{tabular}
\end{center}

As we know the complete situation for $k=3$, here is a table of computer results
for $k=4$. The general bound refers to either Theorem \ref{thm:bnd_n_small}
(including improvements due to the fact that the size of the regular intersecting family
and its degree are integers), or to Theorem \ref{thm:char_n_max}.

\begin{center}
\begin{tabular}{lrrrrrr}
 $n$ & $8$ & $9$ & $10$ & $11$ & $12$ & $13$ \\ \midrule
 General Bound & $34$ & $36$ & $35$ & $33$ & $33$ & $13$ \\ \midrule
 Largest & $32$ & $36$ & $20$ & $11$ & $\geq 12$ & $13$ \\ \midrule
 $\delta$ & $16$ & $16$ & $8$ & $4$ & $\geq 4$ & $4$ \\
\end{tabular}
\end{center}

\section{Conclusion}\label{sec5}

All known finite projective planes have a prime power as order.
It is a famous and long-standing open problem to decide whether there exist projective planes
which do not have prime power order. In light of this it is clear that determining
the existence of regular $k$-uniform intersecting families is hard for $n=k^2-k+1$.

There are several other problems for which it seems to be more feasible to
obtain new results. We list some of them below.

The Bruck-Ryser-Chowla theorem \cite{Bruck1949} implies that if there is a projective
plane of order $k-1$ and $k-1$ is congruent $1$ or $2$ modulo $4$, then $k-1$
is the sum of two squares. This implies that Theorem \ref{thm:bnd_on_n} is
not tight for the orders $6, 14, 21, 22, \ldots$. The case of order $10$ was ruled
out separately by computer \cite{Lam1989}.

\begin{question}
  For those $k$ for which the non-existence of a projective plane of order $k-1$
  is known, what is the largest $n$ for which a regular $k$-uniform intersecting
  family exists?
\end{question}

\begin{question}
  Are there examples for which the bound in Theorem \ref{thm:bnd_n_small}
  with $n \geq 2k+1$ and $s=1$ is tight except for $(n,k) = (7,3)$ and $(n,k) = (9,4)$?\footnote{In an earlier version of this document we mistakenly did not specify $s=1$. For general $s$ Adam S. Wagner noticed that there exists a $3$-regular intersecting family which reaches the bound in Theorem \ref{thm:bnd_n_small}, see \cite{Wagner2019}.}
\end{question}

Theorem \ref{thmdf} is based on the analysis of Boolean functions in the Hamming
graph, that means an investigation on how $0$-$1$-vectors can lie in certain
eigenspaces of the adjacency matrix of the Hamming graph. It would be very interesting
to do this investigation directly in the Johnson scheme with more direct algebraic
techniques to obtain a non-asymptotic version of Theorem~\ref{thmmain1}.

\section*{Acknowledgements}
The research of the first author is supported by ERC advanced grant 320924
and he is supported by a postdoctoral fellowship of the Research Foundation - Flanders (FWO).
The research of the second author is supported the grant RNF 16-11-10014.

redpajama_set_name: RedPajamaArXiv


British journalist Adrian Levy pieced together stories of the 26/11 terror attacks on Mumbai — "the fourth largest city in the world under siege for three days on live TV" — in his latest book The Siege, co-written with Cathy Scott-Clark.
Levy, who believes Mumbai saved itself because "ordinary people acted in extraordinary ways", says the question to ask is why Indian intelligence agencies failed to act upon the inputs they had received before the attacks.
Talking about 26/11 mastermind David Headley, Levy said the Pakistan-born American terrorist had exploited everyone around him to get into the Lashkar-e-Taiba, but "interestingly, the Lashkar did not want him".
Levy also believes the 35-year prison sentence handed to Headley, who has been cooperating with the US agencies, is likely to be reduced as he will "probably make a deal".
Besides reconstructing the events that led to the attacks, The Siege also tells the stories of those held hostage in the Taj hotel in Mumbai.
Levy, who has been visiting India over the past 20 years, also believes the country's "political culture cripples the intelligence agencies".


text:
CONTROL-SA/Links Expands Centralized Security Management
Joanna Doyle
CONTROL-SA/Links, the latest offering in Houston-based BMC Software's security management solution, automates security administration and enables security managers to monitor complex operations across the network while implementing security policies from one central point.
Building on the capabilities of CONTROL-SA, BMC's main security administration product, CONTROL-SA/Links allows enterprise security professionals to create event definitions and automated reactions, in addition to managing user access rights.
"When we introduced CONTROL-SA to the market, customers were empowered to manage users at the enterprise levels for the first time," said Danny Moser, program director for BMC's security management solutions. "With CONTROL-SA/Links customers can achieve that same enterprise level management of all security events."
CONTROL-SA/Links boasts a number of new features intended to make security administration both more reliable and more efficient. The new system monitors applications enterprisewide and alerts security administrators via pager or e-mail as soon as a problem is perceived. The security manager can then use a touch-tone phone to prescribe predefined corrective actions in response to the alert.
Another new feature, Rule-based Automation, allows security managers to integrate human resources applications and security administration. For example, CONTROL/SALinks, can initiate a process in CONTROL-SA such as administering all end-user access rights for a new employee, to ensure that he or she immediately has all necessary access to resources and data in order to be as productive as possible the first day of work.
In addition, CONTROL-SA/Links can intercept messages originating from a wide range of platforms, systems and applications throughout the organization, including AS/400. Messages are consolidated into a single log and can produce a variety of reports related to the information produced, to alert security managers to potential area for improvements in security. Log sources include a variety of operating systems, firewalls, Web servers, e-mail, access control systems, user applications and human resources applications, among others. Finally, in keeping with the intent to make integration more practical, specific CONTROL-SA/Agents are designed to supports a variety of platforms, systems or applications. Those entities not supported by existing agents can still be centrally managed by a non-intrusive connection which is also provided by CONTROL-SA.


text:
Q: Extract point with precise coordinate with on a NetCDF file with Python For my project I want to extract data from a netCDF file, my file represent the Ocean Surface Temperature in function of latitude/longitude. I want to select on point with precise coordinate and then do a mean for each months for several years.
import netCDF4 as nc
import numpy as np
import cdo
import os
import pandas as pd

from scipy.io import netcdf

from netCDF4 import Dataset
from netCDF4 import num2date

import matplotlib as mpl
import matplotlib.pylab as plt

from cdo import Cdo
cdo = Cdo()

data = "/Users/bf/Desktop/Stage
tos_Omon_EC-Earth3_historical_r5i1p1f1_gn_185001-201412.nc"

ds = nc.Dataset(data)
infile = "/Users/bf/Desktop/Stage/tos_Omon_EC-Earth3_historical_r5i1p1f1_gn_185001-201412.nc"

#to select years : 1979-2014
#new name for the file with the selected years : tos_1979-2014.nc
cdo.seldate('1979-01-01,2014-12-31', input=infile, output="tos_1979-2014.nc")

#to select longitute and latitute -> no important now
#new name for the file with the selected longitudes/latitudes : tos_1979-2014_2.nc
cdo.sellonlatbox('-26.45,65.03', input="tos_1979-2014.nc", output="tos_1979-2014_2.nc")

#ymonavg : multi-year monthly statistical values average
cdo.ymonavg(input="tos_1979-2014_2.nc", output="tos_1979-2014_mean_2.nc")

I thought of extract my temperature data (to do a mean of the temperature for each months since 1979 to 2014) and to put them in a data frame, to do graphics thereafter (with mathplolib).
But while using CDO tool, I can't select just one point. Maybe I have to use another tool but I don't know how to do it.
Thanks for your help !

A: to extract a specific gridpoint you need to use remapping interpolation, with the nearest neighbor interpolation.  I think with the python interface it would be
cdo.remapnn('lon=-26.45/lat=65.03', input="tos_1979-2014.nc", output="tos_1979-2014_2.nc")

(I think) -
I have a video guide on this topic in case it is helpful:
https://www.youtube.com/watch?v=YdFo8YAbLTw

redpajama_set_name: RedPajamaStackExchange


We believe that the Scottish National Party could and should have taken stronger action against both of the cases we have listed. The Party has adopted the International Definition of Antisemitism.
In November 2012, Dr Monaghan tweeted that he was "sad to see the proud Jewish people, so hideously persecuted by the Nazis, now persecuting the people of #Gaza".
In August 2015, the JC contacted Dr Monaghan about the tweet, following which he apologised, saying he had not intended to cause offence, however he did not apologise for the views he expressed. He deleted the tweet.
Dr Monaghan lost his seat in the 2017 General Election. As of July 2017, it is unknown whether the SNP has ever taken any disciplinary action against him for his 2012 tweet.
In September 2017, we put this matter to Dr Monaghan, but we did not receive a response.
In November 2015, Ms White retweeted a grotesque cartoon of piglets emblazoned with the flags of countries including Israel, Britain and the US, as well as Al-Qaeda, ISIS and Mossad, all suckling from a large sow labelled with a Star of David along with the words "Bank Rothschild".
After her tweet was made public, her spokesman stated: "This tweet was retweeted in error and has been removed from Sandra White's account. Sandra apologises for any offence caused." On 11th November 2015, Ms White apologised "unreservedly" for retweeting the image, saying she had done so unintentionally. On the same date, the leader of the SNP, Nicola Sturgeon, wrote that she had spoken directly to Ms White about the retweet of the "abhorrent" image and that she would not tolerate antisemitism of any kind or at any level. However, she accepted Ms White's explanation that the retweet was accidental. Ms White retained her seat in the 2016 elections to the Scottish Parliament.
In September 2017, we put this matter to Ms White, but we did not receive a response.


text:
Q: How to resize an image before storing using tapestry5 I am using tapestry5 framework, in which I am using uploadedFile component which is returning me a file which is type of an image.
Now I want to resize that image to 20 x 20, irrespective of the type of the image.
Is there any way with Java API or with Tapestry5 framework to achieve this?

A: This is not functionality that is part of Tapestry, or every likely to be.  The Java2D API has more than sufficient support for converting byte streams into images, rescaling them, and storing them back out.

A: Try this:
BufferedImage resizedImage = new BufferedImage(IMG_WIDTH, IMG_HEIGHT, type);
Graphics2D g = resizedImage.createGraphics();
g.drawImage(originalImage, 0, 0, IMG_WIDTH, IMG_HEIGHT, null);
g.dispose();


New Pesto to Try — Dine on In!
I respect your privacy. I promise only to use your email to send you recipes, new cooking classes, and Pop Up Dinner info.
Thank you! I'm so happy that you will be cooking alongside me. Please share your feedback and any adaptations you did on the recipes.
Boy, what a fun day I had the other day when I came home with my market radishes. I recently roasted them and served them with a sesame salt and a serrano foam. This time, I wanted to do something with the radish tops. I love making pesto with all sorts of greens and this time, the radish tops were no different.
Chop off the tops from the radish. Separate the tops from the radish (use a bowl). Cut off the core from the brussel sprouts and add them all into a bowl (rinse both with clean cold water to remove all the grit).
Once the veggies were rinsed and drained, I added them to a scorching hot skillet with a bit of oil to roast on high. Once they were finished (about 10-15 mins; I like 'em chard and black) I tossed with the radish top pesto. Seriously.... Go run! and make this.
It became a staple all week for veggies, wet rubs, and also a sandwich spread. whoah! go and make it.
Take the tops off the radishes and separate.
Take the core off the brussel sprouts and add to large mixing bowl.
Rinse under cold water both the radish and the sprouts.
Heat the skillet to high heat until it smokes.
Add EVOO and add the radish and sprouts to roast under the high heat.
Meanwhile in the food processor (or blender). Add the last ingredients and set aside.
Toss the roasted veggies with 1/2 of the pesto sauce. Keep the other half for another use.


A contemporary hotel with an enviable seafront location in a city waking up to outdoor life and an appreciation of its art and design heritage.
Launceston, Tasmania's second city, presents a charming combination of colonial architecture and greenery. The city's parks are a delight while heading past its borders will reward you with rolling hills that host a thriving wine industry. It's unsurprising, then, that the city is becoming something of a foodie destination with restaurants fuelled by an influx of fresh, local produce from the surrounding farmland. The hotel itself is right on the city's marina with views over the Tamar River.
To fit with the hotel's dry-dock location, its contemporary building draws its shape and inspiration from seagoing ships. You'll have to choose between river view rooms or city views that look back over Launceston to northern Tasmania's mountain ranges. All contemporary rooms feature en suite bathrooms and comfortable furnishings and come with complimentary WiFi access. For longer stays, the larger suites feature fully equipped kitchens, laundry facilities and a separate lounge and dining area. Their real draw, however, are balconies that make the most of the hotel's sweeping vistas.
Begin the day in the Cube Café where you can enjoy a cooked breakfast and coffee as the wharf starts to wake up. For lunch and dinner, the ground floor Mud Bar and Restaurant serves an eclectic mix of international cuisine. Carefully matched with some superb Tasmanian wines, meals can be served overlooking the marina and North Esk River for year-round Tamar Valley views. Take a stroll down the same wharf and you'll reach award-winning cafés, restaurants and bistros, many specialising in Tasmanian seafood.
Sip on a glass of local wine and watch the comings and goings of the seaport dock or simply take in the view from your room. Friendly staff can also advise you as to the sights and pleasures of the surrounding countryside including the bushland and towering cliffs of Cataract Gorge, just a ten-minute walk from the city centre. Guests also have access, for a fee, to the facilities at Health & Fitness World, a three minute's walk away.
This 1830s homestead is set across 150 sprawling acres of countryside in the Meander Valley's gourmet touring region. Formerly the ancestral home of the first Tasmania-born Premier, Sir Richard Dry, Quamby Estate is one of Tasmania's most historically significant properties.


text:
Красицька Катерина Григорівна (, с. Кирилівка, тепер село Шевченкове Звенигородського району Черкаської області — , с. Зелена Діброва, тепер Городищенського р-ну Черкаської обл.) — старша сестра і нянька Тараса Шевченка. З нею Тарас ходив до Мотронинського монастиря, бував у Чигирині. У повісті «Княгиня» поет залишив про сестру теплі спогади.

 року вийшла заміж за Антона Красицького — селянина-кріпака села Зеленої Діброви, в якому Тарас її часто відвідував.

Катерина народила дванадцятеро дітей: Олену, Федору, Феодосія, Соломію, Степана, Семена, Ганну, Марію, Федора, Якима, Максима і Тетяну, з яких живими залишилось лише четверо — Яким, Степан, Максим і Федора.

Сестра поета мала хист до малювання. Шевченко відвідував її 1843 та у вересні 1845. Згадував у листах до брата Микити.

Померла в 44 роки в один рік з чоловіком та наймолодшою дочкою Тетяною під час епідемії холери 1848 року.

Нащадки
 Фотій Степанович Красицький (син Степана) — художник і педагог.
 Дмитро Филимонович Красицький (внук Якима) — письменник.

Вшанування пам'яті
В Зеленій Діброві, на місці, де стояла хата в якій мешкала Катерина, встановлено пам'ятний знак. Також в селі встановлений їй пам'ятник.

Примітки

Джерела
 Шевченко Катерина Григорівна

Посилання

Родина Тараса Шевченка
Померли від холери
Поховані в Городищенському районі

redpajama_set_name: RedPajamaWikipedia


They are just usual cats aren't they?
Attila the Hun, emperor of all of Europe not part of the Roman Empire, started the refugee migration that caused Venice to be built. Its first non-Byzantine ruler* was in the latter 7th century. It's last ruler was about 1800,(a brief revival after Napoleon's day did not catch on).
Shame if the city will be destroyed after all this time by a rising sea level.
* By the latter 7th century, rival native doges appeared who did not owe fealty to the Byzantine empery. Eventually they effected autonomy and independence from the Empire of the East (corrupted by now).


I'm an illustrator and pattern designer from the UK. I run an Etsy shop selling illustrated greeting cards, prints, and accessories made from fabrics printed with my own pattern designs. I love food and animals, so both crop up in a lot of my designs!


My first ever build and she's a beast and a half!
Running ek full acrylic black blocks on the 980s and bitfenix black block on the i7 with dual 60mm ek rads 480 top and 240 bottom.
I would love to see more pictures of this! Parts are excellent!
Very nicely done. Needs more pictures though!
Still no more pictures? :) How is your build running now that you have used it for a while? I'm specially interested in the Mobo and the case, how was it to setup everything? thx and +1 from me for this clean build.
Really clean! Can we have more pictures?
Great part selection. More photos and a longer description would be nice. Also, your missing a power supply and that's kind of important.
Solid looking build, I fell that CPU was most likely a bit of overkill though. Do you have any more pictures? How does the thing game or run the programs you made it for?
Its a phanteks enthoo primo! The Psu is only visible from behind the mobo panel!! Wow you guys are slacking on keeping up with things that case was released in 2013! FLAWLESS !!!! Lol that bottom radiator is so thick That you NEED the industrials or some Deltas to even think of pull only through it ! So clean MORE pics please!
nice case man. +1 from me.i got the luxe in white myself. i like your parts list also. white cases are where its at. you should check mine out if you get a chance.
Looking good, would like to see more pics of your build.
I have been debating on getting those iPPC Noctuas. What do you think of them so far?
Long overdue update incase anyone still wants to know. I had a leak caused by a pinhole defect in the tube from the cpu to top rad (didnt leak until the system got hot one day playing gta5) had to rip out the watercooling so its on air atm. I'm on my 3rd motherboard or I will be once I get my third one from asus. First one VRM died and cooked my proccessor but props to both intel and asus quick and easy rma and I was up and running. that was jan 2015 nov 2015 mobo 2 had 2 ram slots stop working I could have lived with it but a warranty is a warranty so rmaing again atm turn around was within a week of them getting it but shipping around the holidays is a nightmare.
Things I still love - The case is great still solid and looks good if you like big cases that don't look like a spaceship/racecar, upgraded to a 500gb evo 850 love it 980s are still running great overclock well even on air 5820k doesn't even care what I run it takes it like a champ and keeps going. Asus mobo when working its good when it dies their rma service is quick and professional so no complaints.
Things I would change - different pump and normal noctua fans instead of the 2000rpm ones its not super loud but I would like it to be quieter. 980tis just cause if I had the money for them. Might redo the loop in the new year depending on funds.
Power supply sits sideways on the cable management side.


Michael is a freelance photographer, living in New York City.
Photography has always been an integral part of Michael's life. Some of his earliest memories are of crouching around an old wooden crate in his childhood home, scouring through stacks of white-bordered photos, aged and browning: chubby-legged babies in bikinis playing in backyard blow-up pools, shirtless men drinking beers on whitewater rafting trips, families reuniting over barbecues and around picnic tables covered with red checkered-cloth, neighborhood block parties, holiday gatherings, birthday parties.


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Home » Boxed Teas » Celebration Herbals
15 Items. Showing Items 1 thru 9.
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Chamomile Flowers Tea
Chamomile has been in continuous use throughout the world since the time of the ancient Egyptians. The name Chamomile comes from the Greek word for "Ground Apple," which accurately describes the flower's aromatic scent of apples. Long renowned for its unique effects, Chamomile is a widely accepted herb, often used in cosmetics and perfumery.
Dandelion Leaf Tea
Dandelion's first use was recorded in China in the seventh century. In the eleventh century the Arabs promoted its use, and by the sixteenth century it was well established throughout Europe. It was introduced to North America by the early settlers where it now grows wild. Today, Dandelion greens are common in spring salads. The leaves also bear special properties and good flavour as an herbal tea.
Dandelion Root Tea
The use of Dandelion root as an herb was recorded in China in the seventh century. In the eleventh century the Arabs promoted its use. By the sixteenth century, it was well established throughout Europe. Today, Dandelion root is an important medicinal herb and is also used as a caffeine free coffee substitute. Makes a wonderful tea!
Echinacea Blend Tea
Echinacea is native to the prairie regions of North America. The Native Americans used this herb as a snakebite remedy and taught the settlers its many uses.
Echinacea has been the subject of much research and has become a staple in the modern-day herbal medicine chest.
This blend combines the medicinal properties of the two most popular species of Echinacea to create a powerful combination.
Elderberries Tea
The Elder is a legendary tree, considered magical and sacred in folklore and myths. Some herb manuals refer to it as the medicine chest of the country people. The Elder tree produces creamy-white flowers in late spring which give way to bunches of juicy, purplish black berries in early autumn.
Birds love Elderberries and will eat entire crops if given the opportunity, so harvesting must be done quickly. Today, Elderberries are prized for making juice, pies, desserts and, of course, they make a great fruity tea, hot or cold.
Eleuthero Root Tea
Eleuthero is a small shrub native to Northeastern Asia, where it grown abundantly. Formerly known as 'Siberian Ginseng,' the use of this powerful herb in Traditional Chinese Medicine dates back to 190 A.D.
In recent years, it has been introduced and adapted to parts of North America. Eleuthero is one of the most honoured and ancient of all medicinal herbs, and legend knows it as the 'Root of Life.' Although Eleuthero root is a relatively new addition to the Western world, it has quickly gained in popularity due to its medicinal benefits.
Ginger / Peppermint Tea
Ginger has been used in Asia for over 5,000 years, and remains a key ingredient in both traditional Chinese and Indian Ayurvedic medicine. This spicy rhizome is great medicinally and is widely used as an everyday spice.
Peppermint is known for its clean, refreshing flavour and sensation of coolness. It is widely used as a flavouring agent and fragrant oil. Ginger and Peppermint combine to complement each other for a spicy, yet refreshing tea without caffeine.
Our Ginger Root does not contain sulfates.
For centuries Ginger has been used in the Far East, especially China. This aromatic and spicy plant was introduced to the Americas by the Spaniards and is now widely cultivated in southeast Asia. Today, Ginger is used extensively in many cuisines of the world, and is a traditional drink in many Asian cultures. Our Ginger Root does not contain sulfates.
Hibiscus is cultivated throughout the tropics; its flowers are highly prized for making jams, jellies, sauces, and beverages. Lemongrass is a sub-tropical herb known for its strong citrus flavour and aroma. It became a popular edition to tea blends in the 1970s and was quickly accepted in Europe. By blending these tropical flavours together, we have created a tart and fruity tea that is kid friendly and truly wonderful hot or cold.
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Grand Hotel Bernardin is the first and the largest convention hotel in Slovenia, and looking at the quality of the service, it goes beyond 5 stars. Its wonderful location close to the sea and a private beach provide its guests an idyllic experience.
Situated in one of the most beautiful parts of the Slovenian coast, it is suitable for holding various events. The largest hall, Europa, accommodates up to 1,100 participants. Thanks to its multi-purpose design, the Bernardin Congress Centre is ideal for hosting events such as congresses, seminars, workshops, presentations and business meetings.
This high quality hotel providing remarkable services is located in a quiet and natural environment between Piran and Portorose, directly overlooking the sea. The fully - air - conditioned hotel has 241 comfortably furnished rooms with balconies and sea view, of which 10 executive suites, 11 junior suites and 9 superior rooms. Accommodation is available at GH Bernardin and several other Bernardin hotels at special conference prices.
Bernardin hotels are located between Portorose and Piran, two beautiful towns that offers to every visitor a truly authentic cultural remix - influenced by the Northern Mediterranean, combining inspiring modernity and fascinating history.
Portorose, a city with a holiday atmosphere, lies in a leeward bay and has a very pleasant climate due to its proximity to the sea. The neighbouring, ancient city of Piran is one of the most beautiful Mediterranean cities and one of the best preserved historic cities in Europe.
The whole area is a string of natural attractions and cultural sights, all lined up along only a few kilometres.
Portorose, (literally "Port of Roses"), is an Adriatic - Mediterranean coastal settlement in the Municipality of Piran in south-western Slovenia. Its modern development began in the late 19th century with appearance of first health resorts. In the early 20th century it became one of the grandest seaside resorts in Europe, along with Abbazia, Lido and Grado, then as part of the Austrian Littoral. It is now one of the country's major tourist areas.
Portorose, with its sub-Mediterranean climate, is one of the warmest areas in Slovenia. There are many sunny days. The yearly average amounts to 2,346 hours.
Piran is one of the three major cities in Slovenian Istria. With its medieval architecture and picturesque streets, it is one of the most photogenic cities in the Mediterranean. The old town centre, which is still perfectly preserved, is characterized by the stony narrow streets, compact houses, historical and religious buildings, beautiful Venetian palaces, lively squares, numerous museums and galleries, ancient town walls, its recently reopened aquarium and delicious restaurants. The city is very proud of its various cultural and art establishments such as the Tartini theatre, coastal galleries, Maritime museum and Aquarium.
Much more useful and interesting information about these two cities can be found here.


text:
Q: How to save window contents as an image I want to capture the contents of a JPanel as an image so that I can store it exactly as the user sees it.  Is there a relatively simple way to do this with swing?

A: Use Robot and public BufferedImage createScreenCapture(Rectangle screenRect). and here is the link how to use it

A: From here http://forums.oracle.com/forums/thread.jspa?threadID=2156176&tstart=0
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.image.BufferedImage;
import java.io.File;
import javax.imageio.ImageIO;
import javax.swing.JFrame;
import javax.swing.JPanel;

public class SavePaint extends JPanel
{

    public SavePaint()
    {
        JFrame frame = new JFrame("TheFrame");
        frame.add(this);
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        frame.setSize(400,400);
        frame.setVisible(true);

        try
        {
            BufferedImage image = new BufferedImage(getWidth(), getHeight(), BufferedImage.TYPE_INT_RGB);
            Graphics2D graphics2D = image.createGraphics();
            frame.paint(graphics2D);
            ImageIO.write(image,"jpeg", new File("/home/deniz/Desktop/jmemPractice.jpeg"));
        }
        catch(Exception exception)
        {
            //code
        }
    }

    protected void paintComponent(Graphics g)
    {
        g.drawRect(50,50,50,50);
    }

    public static void main(String[] args)
    {
        new SavePaint();
    }
}

A: use new java.awt.Robot().createScreenCapture(screenRect)


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Robert Starke, MD can be found at 1321 Northwest 14th Street . The following is offered: Medical Consultants . In Miami there are 65 other Medical Consultants. An overview can be found here.


Volunteer Missionary Candidates to the Pendjari of Benin since 2018.
Chris grew up at a Seventh-day Adventist church camp in Tennessee. Over the years of working at camp, he has developed a wide variety of hands on skills. Meanwhile, Erin grew up in a small goal-mining town in Western Kentucky. Frequently she visited the church camp where Chris lived. Our first memories of interacting with each other are when we were about 8 years old. Over the years we went different directions and it wasn't until 20 years after our first meeting that the Lord brought us together and we were married. Almost 6 years ago the Lord brought us back to the camp where Chris grew up and what a blessing it has been to be sharing Jesus through camp ministry.
We have accepted the call to the Pendjari region of Benin, West Africa. We will live in the local community, getting to know the people and developing friendships with them. We hope to show our love for them by creating job skills training opportunities and job industries. Our prayer is that we will be able to lead our new friends into relationships with Jesus and provide support and mentoring to them as they grow in Him.
As we have reflected over the past 12 months, we cannot even begin to imagine how the past year's blessings and sorrows will continue to impact us and our growing Pendjari family. Truly, "…Eye hath not seen, nor ear heard, neither have entered into the heart of man the things which God hath prepared for them that love him" (1 Co. 2:9).
Commissioning Service with traditional Beninese food being served at potluck after the service. AFTER POTLUCK WE WILL BE SHOWING AN AFM VIDEO FOCUSED ON A BENIN PROJECT WITH TIME FOR QUESTIONS AND ANSWERS TO FOLLOW.


" Together we may give our children the roots to grow and the wings to fly. "
At Rims we believe that we offer the best teaching and learning experience that have been specially planned to meet the need of present young people.
The RIMS International School under the aegis of RIMS Foundation is a centre of excellence offering systematic and innovative learning strategies with a unique combination of both Indian and International perspectives. A dynamic team of professionals from Kannur who are parents founded the RIMS Foundation in May 2009 under the chairmanship of Dr.P.A.Ibrahim , a highly acclaimed visionary, generous philanthropist, a renowned entrepreneur and industrialist both in India and the overseas. He presently heads many educational institutions and business ventures both in India and abroad..
The school places special emphasis on developing the dormant imagination and creativity of children.The school is divided into four Houses .A house mentor is responsible in coaching each house.The house leaders motivate the students to perform in activities which need both physical stamina and mental stability.
All children need acknowledgement and appreciation for their feelings and creativity.So allowing children to express their souls through art would be a holistic experience for both the creator and the admirer.It goes without saying that the bulletin boards of the school give ample opportunities for the children to showcase their imagination and dexterity.Good Art classes by resourse persons, art competition and exhibitions are common activities in the school to develop their spacial intelligence.
The Rims International School takes pride in its Chithrajali Arts and Craft club.The school conduct many drawing and coloring competetion every year.
The rimians are always encouraged and awarded prizes for putting up their best works.
Undoubtedly sports activities are inevitable in shaping the lives of the young.The children here knows that wanting to win is more important than winning.Tthe score never discourages their spirits.the community , the school administrators and the mentors make sure that they shape their future athelete from the school sports grounds .Since sports activities are hundrfed fold better when compared to other entertainmants like television ,radio,internet.They are held at the school grounds yearly with a goor preparation to instill not onlyphysical vigour but also good characters like disipline,responsibility , self confidence,sacrifiece and accountability in students.
Rimians justify the saying that' a healthy mind is found in a healthy mind is found in a healthy body. The school celebrates sports events on a yearly basis. Important people in the sports field are invited to inaugurate and educate children during sports events. Each houses have different flags, students of RIMS International School are given innumerable opportunities to play a variety of sports. The children are facilitated with the necessary equipment to play sports from team games to individual sports . Even Karate classes are given here for students by eminent coaches irrespective of them gentre. The play zone in front of the school proclaim the important given for plays. Generally all children enjoy playing game like frog jumping, running, potatoes gathering, sack race and so on. Book balancing, threading the needle , lemon spoon race , relay race etc are the other games that entertain the audience. Rimian are adept at mass drills and boys are good at various geometric displays . Identifying the winners, acknowledging them and awarding them prizes go hand in hand . The school in proud to announce about its swimming pool which world he completed very soon.
It gives me immense pleasure to learn that the RIMS had made a remarkable identity with a stupendous growth and development. I appreciate the principal and the staff for their single minded devotion and dedication for changing the noble vision of life oriented education in to a living reality. Insha Allah. The RIMS International School will be soon ranked as one of the top schools in our country for its unique features. May the Almighty lead the RIMS and the Rimians to the epitome of success.
Welcome to the Rims International School Website . As you read through the pages of this site, you will learn about the school, students and get a feel of the vibrancy and strength of our school community.One of the most important decisions that any family makes is the school they choose for their children's education. At Rims we believe that we offer the best teaching and learning experience that have been specially planned to meet the need of present young people.Our school Rims International is committed to give its students a strong foundation in education by creating a nurturing environment to make learning a pleasurable experience .Our school is also committed to creating an environment to enable the children of all ages to fulfill their potential in all spheres of life .Our curriculum is broad and balanced and provides for the academic, moral, physical, creative, spiritual and social development of every child.
At Rims International School we aim not only to peruse academic excellence but also to motivate and inspire our students to be lifelong learners critical thinkers and productive members of the global society.Our school believes that developing a good sense of discipline and moral values among students will enable them to be a good human being in the society.Our education is centered in appreciation encouragement enthusiasm and affection rather than criticism and punishment.I am confident that our school will always be a beacon of light, guiding the destiny of our students and fulfilling of the social motto 'Skills for life'.
Working Days For Pre- Primary, Primary & Grades I - V - Monday to Friday.
Office Timings: 09.00 a.m. to 04.00p.m. (Monday to Friday) Saturdays till 12.00 p.m only.
Lunch Break : 01:15 p.m. to 02.00.p.m.
Principal/Sr Principal : 03:00 p.m. to 04:30 p.m.
The Principal/Sr Principal & other officers in key position shall be available to the parents for consultation only by prior appointment during the visiting hours.
It is mandatory for all pupils to wear the school uniform. All students must be in full and correct uniform. We ask for parental support in achieving the best possible standard for our students.
Polo Frock (Orange ,Green & Yellow) with knee length pants(Navy Blue).
Polo T-Shirt (Orange,Green & Yellow) with Navy Blue Bermudas.
Wednesdays : White full sleeved salwar with House color Jacket(Sports Wear).
Regular days : Light Brown Full pants & Light Brown striped Full sleeved shirt.
Wednesdays : House color Polo T-shirt & Track pants (Sports Wear).
Black shoes and White socks with blue stripes.
Regular days : Light Brown shoes and light brown socks with gold colour bands.
Wednesdays : White shoes and White socks.
Note : Jewelery is not allowed other than a watch (5th upwards only) and a pair of stud ear rings(girls). Do not use hair gel, hair cream etc. Hats can be brought to school for use when out in the sun.Girls must tie their hair on both sides with light brown satin ribbons. Brown hair bands with regular uniform and white hair band on Wednesdays are to be used for short hair.Muslim girls from Grade III onwards can cover their hair with a scarf light brown on regular days and white scarf on Wednesdays.
It gives me immense pleasure to learn that the RIMS had made a remarkable identity with a stupendous growth and development. I appreciate the principal and the staff for their single minded devotion... Read More..
The school places special emphasis on developing the dormant imagination and creativity of children.The school is divided into four Houses .A house mentor is responsible in coaching... Read More..
Working Days: For Pre- Primary, Primary & Grades I - V - Monday to Friday.
For Grades VI - XII - Monday to Saturday Working Hours: Grades I to Above : 09:10am to 03:45pm Read More..