text:
\section{Introduction}

Let $R$ be a ring and $A \subset R$.  The \emph{sumset} of $A$ is the set $A + A = \{ a + b : a ,b \in A \}$,
and the \emph{product set} of $A$ is the set $A \cdot A = \{ a \cdot b : a , b \in A \}$.
A well-studied problem in arithmetic combinatorics is to prove non-trivial lower bounds on the quantity
\[
\max \{ | A + A | , |A \cdot A | \}
\]
under suitable hypothesis on $R$ and $A$.  One of the first results of this type is due to Erd\H{o}s
and Szemer\'{e}di \cite{es}.  They proved that if $R = \mathbb{Z}$ and $A$ is finite, then there are positive constants $c$ and $\epsilon$, both independent of $A$, such that
\[
\max \{ |A + A| , |A \cdot A | \} \geq c |A|^{1 + \epsilon}.
\]
This improves the trivial lower bound of $\max \{ |A + A| , |A \cdot A | \} \geq |A|$.  Erd\H{o}s and Szemer\'{e}di conjectured that the correct exponent is $2 - o(1)$ where $o(1) \rightarrow 0$ as $|A| \rightarrow \infty$.
Despite a significant amount of research on this problem, this conjecture is still open.  For some time the best known exponent was $4/3 - o(1)$ due to Solymosi \cite{soly} who proved that for
any finite set $A \subset \mathbb{R}$,
\[
\max \{ |A + A | , |A \cdot A | \} \geq \frac{ |A|^{4/3} }{ 2 (  \log |A| )^{1/3} }.
\]
Very recently, Konyagin and Shkredov \cite{ks} announced an improvement of the exponent to $4/3 + c- o(1)$ for any
$c < \frac{1}{20598}$.

Another case that has received attention is when $R$ is a finite field.  Let $p$ be a prime and
let $A \subset \mathbb{Z}_p$.  Bourgain, Katz, and Tao \cite{bkt} proved that if $p^{ \delta } < |A| < p^{1- \delta}$
where $0 < \delta < 1/2$, then
\begin{equation}\label{eq:bkt}
\max \{ |A + A| , |A \cdot A| \} \geq c  |A|^{1 + \epsilon }
\end{equation}
for some positive constants $c$ and $\epsilon$ depending only on $\delta$.  Hart, Iosevich, and Solymosi \cite{his} obtained bounds that give an explicit dependence of $\epsilon$ on $\delta$.
Let $q$ be a power of an odd prime, $\mathbb{F}_q$ be the finite field with $q$ elements, and $A \subset \mathbb{F}_q$.
In \cite{his}, it is shown that if
$|A + A| = m$ and $|A \cdot A | = n$, then
\begin{equation}\label{eq:his}
|A|^3 \leq \frac{ c m^2 n |A| }{ q} + c q^{1/2} mn
\end{equation}
where $c$ is some positive constant.  Inequality (\ref{eq:his}) implies a non-trival sum-product estimate when
$q^{1/2} \ll |A| \ll q $.  We write $f \ll g$ if $f = o(g)$.  Using a graph theoretic approach, the fourth author \cite{vinh} and Vu \cite{vu} improved (\ref{eq:his}) and as a result, obtained a better sum-product estimate.

\begin{theorem}[\cite{vinh}]\label{vinh th}
Let $q$ be a power of an odd prime.  If $A \subset \mathbb{F}_q$, $|A+A| = m$ and $|A \cdot A| = n$, then
\[
|A|^2 \leq \frac{mn|A|}{q} + q^{1/2} \sqrt{ m n } .
\]
\end{theorem}

\begin{corollary}[\cite{vinh}]\label{vinh cor}
If $q$ is a power of an odd prime and $A \subset \mathbb{F}_q$, then there is a positive constant $c$ such that the following hold.
If $q^{1/2} \ll |A| < q^{2/3}$, then
\[
\max \{ |A + A | , |A \cdot A | \} \geq \frac{ c |A|^2 }{q^{1/2}} .
\]
If $q^{2/3} \leq |A| \ll q$, then
\[
\max \{ |A + A | , |A \cdot A | \} \geq  c ( q |A| )^{1/2} .
\]
\end{corollary}

In the case that $q$ is a prime, Corollary \ref{vinh cor} was proved by Garaev \cite{garaev} using exponential sums.
 Cilleruelo \cite{cil} also proved related results using dense Sidon sets in finite groups involving $\mathbb{F}_q$ and $\mathbb{F}_q^*$.
In particular, versions of Theorem \ref{vinh th2} and (\ref{gyarmati}) (see below) are proved in \cite{cil}, as well as several other results concerning equations in $\mathbb{F}_q$ and sum-product estimates.

Theorem \ref{vinh th} was proved using the following Szemer\'{e}di-Trotter type theorem in $\mathbb{F}_q$.
\begin{theorem}[\cite{vinh}]\label{vinh th2}
Let $q$ be a power of an odd prime.  If $P$ is a set of points and $L$ is a set of lines in $\mathbb{F}_q^2$, then
\[
| \{ (p , l ) \in P \times L : p \in l \} | \leq \frac{ |P| |L| }{q} + q^{1/2} \sqrt{ |P| |L| }.
\]
\end{theorem}
We remark that a Szemer\'{e}di-Trotter type theorem in $\mathbb{Z}_p$ was obtained in \cite{bkt} using the sum-product estimate (\ref{eq:bkt}).

In this paper, we generalize Theorem \ref{vinh th}, Corollary \ref{vinh cor}, and Theorem \ref{vinh th2} to finite quasifields.
Any finite field is a quasifield.  There are many examples of quasifields which are not fields; see for example, Chapter 5 of \cite{demb} or Chapter 9 of \cite{hp}.  Quasifields appear extensively in the theory of projective planes.

\begin{theorem}\label{main th}
Let $Q$ be a finite quasifield with $q$ elements.  If $A \subset Q \backslash \{0 \}$, $|A+A| = m$ and $|A \cdot A | = n$, then
\[
|A|^2 \leq \frac{mn |A| }{q} + q^{1/2}\sqrt{mn}.
\]
\end{theorem}
\medskip

Theorem \ref{main th} gives the following sum-product estimate.

\begin{corollary}\label{main cor}
Let $Q$ be a finite quasifield with $q$ elements and $A \subset Q \backslash \{0 \}$.  There is a positive constant $c$ such that the following hold.

If $q^{1/2} \ll |A| < q^{2/3}$, then
\[
\max \{ |A + A | , |A \cdot A | \} \geq c  \frac{ |A|^2 }{ q^{1/2} } .
\]

 If $q^{2/3} \leq |A| \ll q$, then
\[
\max \{ |A + A| , |A \cdot A| \} \geq c ( q |A| )^{1/2}.
\]
\end{corollary}

From Corollary \ref{main cor} we conclude that any algebraic object that is rich enough to coordinatize a projective plane must satisfy a non-trivial sum-product estimate. Following \cite{vinh}, we prove a Szemer\'{e}di-Trotter type theorem and then use it to deduce Theorem \ref{main th}.

\begin{theorem}\label{main th2}
Let $Q$ be a finite quasifield with $q$ elements.   If $P$ is a set of points and $L$ is a set of lines in $Q^2$, then
\[
| \{ (p , l ) \in P \times L : p \in l \} | \leq \frac{ |P| |L| }{q} + q^{1/2} \sqrt{ |P| |L| }.
\]
\end{theorem}

Another consequence of Theorem \ref{main th2} is the following corollary.

\begin{corollary}\label{v2:cor 5}
If $Q$ is a finite quasifield with $q$ elements and $A \subset Q$, then there is a positive constant $c$ such that
\[
| A \cdot (A + A ) | \geq c\min \left\{ q , \frac{ |A|^3 }{q} \right\}.
\]
Furthermore, if $|A|\gg q^{2/3}$, then one may take $c = 1+o(1)$.
\end{corollary}

The next result generalizes Theorem 1.1 from \cite{vinh 3 var}.

\begin{theorem}\label{v2:theorem 4}
Let $Q$ be a finite quasifield with $q$ elements.  If $A , B , C \subset Q$, then
\[
| A + B \cdot C | \geq q - \frac{q^3}{|A||B||C| + q^2}
\]
\end{theorem}

Our methods in proving the above results can be used to generalize theorems concerning the solvability of equations over finite fields.  Let $p$ be a prime and let $A , B , C , D \subset \mathbb{Z}_p$.  S\'{a}rk\"{o}zy \cite{sar} proved that
if $N(A,B,C,D)$ is the number of solutions to $a + b  = cd$ with $(a,b,c,d) \in A \times B \times C \times D$, then
\begin{equation}\label{gyarmati}
\left| N(A,B,C,D) - \frac{ |A| |B| |C| |D| }{p} \right| \leq p^{1/2} \sqrt{ |A| |B| |C| |D| }.
\end{equation}
In particular, if $|A| |B| |C| |D| > p^3$, then there is
an $(a,b,c,d) \in A \times B \times C \times D$ such that $a+b = cd$.
This is best possible up to a constant factor (see \cite{sar}).  It was generalized to finite fields of odd prime power order by
Gyarmati and S\'{a}rk\"{o}zy \cite{gs}, and then by the fourth author \cite{vinh2} to systems of equations over $\mathbb{F}_q$.  Here we generalize the result of Gyarmati and S\'{a}rk\"{o}zy to finite quasifields.

\begin{theorem}\label{main th3}
Let $Q$ be a finite quasifield with $q$ elements and let $A , B , C, D \subset Q$.  If $\gamma \in Q$ and
$N_{ \gamma } (A,B,C,D)$ is the number of solutions to
$a + b + \gamma = c \cdot d$ with $a \in A$, $b \in B$, $c \in C$, and $d \in D$, then
\[
\left| N_{ \gamma} (A , B , C, D)  - \frac{ (q+1)|A| |B| |C| |D| }{q^2 + q + 1} \right| \leq q^{1/2} \sqrt{ |A| |B| |C| |D| }.
\]
\end{theorem}

Theorem \ref{main th3} implies the following Corollary which generalizes Corollary 3.5 in \cite{vinh3}.

\begin{corollary}\label{main cor3}
If $Q$ is a finite quasifield with $q$ elements and $A, B , C , D \subset Q$ with $|A||B||C||D| > q^{3}$, then
\[
Q  = A + B + C \cdot D.
\]
\end{corollary}

We also prove a higher dimensional version of Theorem \ref{main th3}.

\begin{theorem}\label{v2:theorem 1}
Let $d \geq 1$ be an integer.
If $Q$ is a finite quasifield with $q$ elements and $A \subset Q$ with $|A| > q^{ \frac{d+2}{2d+2} }$,
then
\[
Q  = A +A + \underbrace{A \cdot A + \dots + A \cdot A}_{d~\textup{terms}}.
\]
\end{theorem}

Another problem considered by S\'{a}rk\"{o}zy was the solvability of the equation $ab + 1 = cd$ over $\mathbb{Z}_p$.   S\'{a}rk\"{o}zy \cite{sar2} proved a result in $\mathbb{Z}_p$ which was later generalized to the finite field setting in
\cite{gs}.

\begin{theorem}[Gyarmati,  S\'{a}rk\"{o}zy]\label{gs:th2}
Let $q$ be a power of a prime and $A , B , C , D \subset \mathbb{F}_q$.  If $N(A , B , C , D)$ is the number of solutions to $ab + 1 = cd$ with $a \in A$, $b \in B$, $c \in C$, and $d \in D$, then
\[
\left| N (A,B,C,D) - \frac{ |A| |B| |C| |D| }{q} \right| \leq 8 q^{1/2} ( |A| |B| |C| |D| )^{1/2}  + 4q^2.
\]
\end{theorem}

Our generalization to quasifields is as follows.

\begin{theorem}\label{v2:theorem 2}
Let $Q$ be a finite quasifield with $q$ elements and kernel $K$.  Let $\gamma \in Q \backslash \{0 \}$, and $A , B , C , D \subset Q$.  If $N_{ \gamma} (A,B,C,D)$ is the number of solutions to $a \cdot b + c \cdot d = \gamma$, then
\[
\left| N_{ \gamma} (A,B,C,D) - \frac{ |A| |B| |C| |D| }{q} \right| \leq  q  \left( \frac{ |A| |B| |C| |D| }{ |K| - 1} \right)^{1/2}.
\]
\end{theorem}

\begin{corollary}
Let $Q$ be a quasifield with $q$ elements whose kernel is $K$.  If $A , B  , C , D \subset Q$ and
$|A| |B| |C| |D| > q^4 ( |K| - 1)^{-1}$, then
\[
Q \backslash \{ 0 \} \subset A \cdot B + C \cdot D.
\]
\end{corollary}

By appropriately modifying the argument used to prove Theorem \ref{v2:theorem 2}, we can prove a higher dimensional version.

\begin{theorem}\label{v2:theorem 3}
Let $Q$ be a finite quasifield with $q$ elements whose kernel is $K$.
If $A \subset Q$ and $|A| > q^{ \frac{1}{2} + \frac{1}{d} } ( |K|  - 1)^{-1/2d}$, then
\[
Q\backslash \{ 0 \} \subset \underbrace{A \cdot A + \dots + A \cdot A}_{d~\textup{terms}}.
\]
\end{theorem}

If $Q$ is a finite field, then $|K| = q$, and the bounds of Theorems \ref{v2:theorem 2} and \ref{v2:theorem 3} match the bounds obtained by Hart and Iosevich in \cite{hi}.

The rest of the paper is organized as follows.  In Section 2 we collect some preliminary results.
Section 3 contains the proof of Theorem \ref{main th}, \ref{main th2}, and \ref{main th3}, as well
as Corollary \ref{main cor}, \ref{v2:cor 5}, and \ref{main cor3}.
Section 4 contains the proof of Theorems \ref{v2:theorem 4}, \ref{v2:theorem 1},
\ref{v2:theorem 2}, and \ref{v2:theorem 3}.

\section{Preliminaries}

We begin this section by recalling the definition of a quasifield.
A set $L$ with a binary operation $\cdot$ is called a \emph{loop} if
\begin{enumerate}
\item the equation $a \cdot x = b$ has a unique solution in $x$ for every $a,b \in L$,
\item the equation $y \cdot a = b$ has a unique solution in $y$ for every $a,b \in L$, and
\item there is an element $e \in L$ such that $e \cdot x = x \cdot e = x$ for all $x \in L$.
\end{enumerate}
A \emph{(left) quasifield} $Q$ is a set with two binary operations $+$ and $\cdot$ such that $(Q , + )$ is a group with additive identity 0, $(Q^* , \cdot )$ is a loop where $Q^* = Q \backslash \{0 \}$, and the following three conditions hold:
\begin{enumerate}
\item $a \cdot (b + c) = a \cdot b + a \cdot c$ for all $a,b,c \in Q$,
\item $0 \cdot x = 0$ for all $x \in Q$, and
\item the equation $a \cdot x = b \cdot x + c$ has exactly one solution for every $a,b,c \in Q$ with $a \neq b$.
\end{enumerate}

We use 1 to denote the identity in the loop $(Q^* , \cdot )$.  It is a consequence of the definition that $(Q , +)$ must be an abelian group.
One also has $x \cdot 0 = 0$ and $x \cdot (-y) = - ( x \cdot y )$ for all $x,y \in Q$ (see \cite{hp}, Lemma 7.1).
For more on quasifields, see Chapter 9 of \cite{hp}.  A \emph{(right) quasifield} is required to satisfy the right distributive law instead of the left distributive law.  The \emph{kernel} $K$ of a quasifield $Q$ is the set of all elements $k \in Q$ that satisfy
\begin{enumerate}
\item $(x + y ) \cdot k = x \cdot k + y \cdot k$ for all $x,y \in Q$, and
\item $(x \cdot y) \cdot k = x \cdot ( y \cdot k )$ for all $x,y \in Q$.
\end{enumerate}
Note that $(K , +)$ is an abelian subgroup of $(Q , +)$ and $(K^* , \cdot )$ is a group.

\begin{lemma}\label{v2:l0}
If $a \in Q$ and $\lambda \in K$, then $-( a \cdot \lambda) = ( -a) \cdot \lambda$.
\end{lemma}
\begin{proof}
First we show that $a \cdot (-1) = -a$.  Indeed, $a \cdot (1 + (-1) ) = a \cdot 0 = 0$ and so $a + a \cdot (-1) =0$.  We conclude that $-a  = a \cdot (-1)$.  If $\lambda \in K$, then
\begin{eqnarray*}
- ( a \cdot \lambda ) & =&  a \cdot ( - \lambda ) = a \cdot ( 0 - \lambda ) = a \cdot ( (0 - 1) \cdot \lambda ) \\
& =& (a \cdot (0 - 1) ) \cdot \lambda = ( 0 + a \cdot (-1) ) \cdot \lambda  = ( -a ) \cdot \lambda.
\end{eqnarray*}
\end{proof}

For the rest of this section, we assume that $Q$ is a finite quasifield with $|Q| = q$.
We can construct a projective plane $\Pi = ( \mathcal{P} , \mathcal{L} , \mathcal{I} )$ that is coordinatized by $Q$. We will follow the notation of \cite{hp} and refer the reader to Chapter 5 of \cite{hp} for more details.  Let $\infty$ be a symbol not in $Q$.  The points of $\Pi$ are
\[
\mathcal{P} = \{(x,y) : x,y \in Q \} \cup \{ (x) : x \in Q \} \cup \{ ( \infty) \}.
\]
The lines of $\Pi$ are
\[
\mathcal{L} = \{ [m,k] : m,k \in Q \} \cup \{ [m] :  m \in Q \} \cup \{ [\infty] \}.
\]
The incidence relation $\mathcal{I}$ is defined according to the following rules:
\begin{enumerate}
\item $(x,y) \mathcal{I} [m,k]$ if and only if $m \cdot x + y = k$,
\item $(x,y) \mathcal{I} [k]$ if and only if $x = k$,
\item $(x) \mathcal{I} [m,k]$ if and only if $x = m$,
\item $(x) \mathcal{I} [ \infty ]$ for all $x \in Q$, $( \infty ) \mathcal{I} [k]$ for all $k \in Q$, and
$( \infty ) \mathcal{I} [ \infty ]$.
\end{enumerate}

Since $|Q| = q$, the plane $\Pi$ has order $q$.

Next we associate a graph to the plane $\Pi$.  Let $\mathcal{G} ( \Pi )$ be the bipartite graph with parts $\mathcal{P}$ and $\mathcal{L}$ where $p \in \mathcal{P}$ is adjacent to $l \in \mathcal{L}$ if and only if $p \mathcal{I} l$ in $\Pi$.  The first lemma is known
(see \cite{bcn}, page 432).

\begin{lemma}\label{l1}
The graph $\mathcal{G}( \Pi )$ has eigenvalues $q + 1$ and $-(q+1)$, each with multiplicity one.  All other eigenvalues of $\mathcal{G}( \Pi )$ are $ \pm q^{1/2}$.
\end{lemma}

The next lemma is a bipartite version of the well-known Expander Mixing Lemma.

\begin{lemma}[Bipartite Expander Mixing Lemma]\label{bipartite expander mixing lemma}
Let $G$ be a $d$-regular bipartite graph on $2n$ vertices with parts $X$ and $Y$.  Let $M$ be the adjacency matrix of $G$.
Let $d=\lambda_1\geq \lambda_2 \geq \cdots \geq \lambda_{2n} = -d$ be the eigenvalues of $M$ and define $\lambda = \max_{i\not=1,2n} |\lambda_i|$. If $S\subset X$ and $T\subset Y$, then
\[
\left| e(S,T) - \frac{d|S||T|}{n}\right| \leq \lambda\sqrt{|S||T|}.
\]
\end{lemma}
\begin{proof}
Assume that the columns of $M$ have been been ordered so that the columns corresponding to the vertices of $X$ come before the columns corresponding to the vertices of $Y$.
For a subset $B\subset V(G)$, let $\chi_B$ be the characteristic vector for $B$. Let $\{x_1, \dots ,x_{2n}\}$ be an orthonormal set of eigenvectors for $M$. Note that since $G$ is a $d$-regular bipartite graph, we have
\begin{align}
& x_1 = \frac{1}{\sqrt{2n}}\left( \chi_X + \chi_Y\right), \label{first eigenvector}\\
& x_{2n} = \frac{1}{\sqrt{2n}} \left(\chi_X - \chi_Y\right) \label{last eigenvector}.
\end{align}
Now $\chi_S^T M \chi_T = e(S,T)$. Expanding $\chi_S$ and $\chi_T$ as linear combinations of eigenvectors yields
\begin{align*}
 e(S,T) = \left(\sum_{i=1}^{2n} \langle \chi_S, x_i\rangle x_i \right)^T M \left( \sum_{i=1}^{2n} \langle \chi_T, x_i\rangle x_i \right)
 = \sum_{i=1}^{2n} \langle \chi_S, x_i\rangle \langle \chi_T, x_i\rangle \lambda_i.
\end{align*}
Now by \eqref{first eigenvector} and  \eqref{last eigenvector},
$\langle \chi_S, x_1\rangle = \langle \chi_S, x_{2n}\rangle  = \frac{1}{\sqrt{2n}} |S|$ and $\langle \chi_T, x_1\rangle = - \langle \chi_T, x_{2n}\rangle = \frac{1}{\sqrt{2n}} |T|$. Since $\lambda_1 = -\lambda_{2n} = d$, we have
\begin{align*}
\left| e(S,T)-\frac{2d|S||T|}{2n}\right| &= \left|\sum_{i=2}^{2n-1} \langle \chi_S, x_i\rangle \langle \chi_T, x_i\rangle \lambda_i \right|\\ &\leq  \lambda \sum_{i=2}^{2n-1} \left| \langle \chi_S, x_i \rangle \langle \chi_T, x_i\rangle \right|\\
& \leq \lambda \left(\sum_{i=2}^{2n-1} \langle \chi_S, x_i\rangle^2\right)^{1/2} \left(\sum_{i=2}^{2n-1} \langle \chi_T, x_i\rangle^2 \right)^{1/2} & \mbox{(by Cauchy-Schwarz).}
\end{align*}
Finally by the Pythagorean Theorem,
\[
\sum_{i=2}^{2n-1} \langle \chi_S, x_i\rangle^2 = |S| - \frac{2|S|^2}{2n} < |S|
\]
and
\[
\sum_{i=2}^{2n-1} \langle \chi_T, x_i\rangle^2 = |T| - \frac{2|T|^2}{2n} < |T|.
\]
\end{proof}

\bigskip

Combining Lemmas \ref{l1} and \ref{bipartite expander mixing lemma} gives the next lemma.

\begin{lemma}\label{beml}
For any $S \subset \mathcal{P}$ and $T \subset \mathcal{L}$,
\begin{equation*}
\left| e(S , T) - \frac{ (q + 1) |S| |T| }{ q^2 + q + 1} \right| \leq  q^{1/2}\sqrt{|S||T|}
\end{equation*}
where $e(S,T)$ is the number of edges in $ \mathcal{G} ( \Pi )$ with one endpoint in $S$ and the other in $T$.
\end{lemma}

We now state precisely what we mean by a line in $Q^2$.

\begin{definition}
Given $a,b \in Q$, a line in
$Q^2$ is a set of the form
\[
 \{ ( x , y ) \in Q^2 : y = b \cdot x + a \} ~~ \mbox{or} ~~ \{ (a,y) : y \in Q \}.
\]
\end{definition}

When multiplication is commutative, $b \cdot x + a = x \cdot b  + a$.  In general, the binary operation $\cdot$ need not be commutative and so we write our lines with the slope on the left.

The next lemma is due to Elekes \cite{elekes} (see also \cite{tv}, page 315).  In working in a (left) quasifield, which is not required to satisfy the right distributive law, some care must be taken with algebraic manipulations.

\begin{lemma}\label{l3}
Let $A \subset Q^*$.  There is a set $P$ of $|A +  A| |A \cdot A |$ points and a set
$L$ of $|A|^2$ lines in $Q^2$ such that there are at least $|A|^3$ incidences between $P$ and $L$.
\end{lemma}
\begin{proof}
Let $P = (A + A ) \times ( A \cdot A)$ and
\[
l(a , b) = \{ (x,y) \in Q^2 : y = b \cdot x - b \cdot a \}.
\]
Let $L = \{ l(a,b) : a , b \in A \}$.  The statement that $|P| = |A + A| | A \cdot A|$ is clear from the definition of $P$.
Suppose $l(a,b)$ and $l(c,d)$ are elements of $L$ and $l(a,b)  = l(c,d)$.
We claim that $(a,b)  = (c,d)$.  In a quasifield, one has $x \cdot 0 = 0$ for every $x$, and $x \cdot (-y) = - (x \cdot y)$ for every $x$ and $y$ (\cite{hp}, Lemma 7.1).
The line $l(a,b)$ contains the points $(0, -b \cdot a)$ and
$(1 , b - b \cdot a)$.  Furthermore, these are the unique points in $l(a,b)$ with first coordinate 0 and 1, respectively.  Similarly,
the line $l(c,d)$ contains the points $(0 , - d \cdot c)$ and $(1 , d - d \cdot c)$.  Since $l(a,b) = l(c,d)$, we must have that
$-b \cdot a = -d \cdot c$ and $b - b \cdot a = d - d \cdot c$.  Thus, $b = d$ and so $b \cdot a = b \cdot c$.  We can rewrite this equation as $b \cdot a - b \cdot c = 0$.  Since $- x \cdot y = x \cdot (-y)$ and $Q$ satisfies the left distributive law, we have $b  \cdot (a - c) = 0$.  If $a = c$, then $(a,b) = (c,d)$ and we are done.  Assume that $a \neq c$ so that $a - c \neq 0$.  Then we must have $b = 0$ for if $b \neq 0$, then the product $b \cdot (a - c)$ would be contained in $Q^*$ as multiplication is a binary operation on $Q^*$.  Since $A \subset Q^*$, we have $b \neq 0$.  It must be the case that $a = c$.  We conclude that each pair $(a,b) \in A^2$ determines a unique line in $L$ and so $|L| = |A|^2$.

Consider a triple $(a,b,c) \in A^3$.  The point $(a+c , b \cdot c)$ belongs to $P$ and is incident to $l(a,b) \in L$ since
\[
b \cdot (a + c) - b \cdot a = b \cdot a + b \cdot c - b \cdot a = b \cdot c.
\]
Each triple in $A^3$ generates an incidence and so there are at least $|A|^3$ incidences between $P$ and $L$.
\end{proof}

\section{Proof of Theorem \ref{main th}, \ref{main th2}, and \ref{main th3}}

Throughout this section, $Q$ is a finite quasifield with $q$ elements,  $\Pi = ( \mathcal{P} , \mathcal{L} , \mathcal{I} )$
is the the projective plane coordinatized by $Q$ as in Section 2.  The graph $ \mathcal{G} ( \Pi )$ is the bipartite graph
defined before Lemma \ref{l1} in Section 2.

\bigskip

\begin{proof}[Proof of Theorem \ref{main th2}]
Let $P \subset Q^2$ be a set of points and view $P$ as a subset of $\mathcal{P}$.  Let $r(a,b) = \{ (x,y) \in Q^2 : y = b \cdot x + a \}$, $R \subset Q^2$, and let
\[
L = \{ r(a,b) : (a,b) \in R \}
\]
be a collection of lines in $Q^2$.  The point $p = (p_1 , p_2)$ in $P$ is incident to the line $r(a,b)$ in $L$ if and only if $p_2 = b \cdot p_1 + a$.  This however is equivalent to $(p_1 , - p_2) \mathcal{I} [ b , -a ]$ in $\Pi$.  If
$S = \{ (p_1 , - p_2) : (p_1 , p_2 ) \in P \}$ and $T = \{ [ b , -a ] : ( a,b) \in R \}$, then
\[
| \{ (p , l ) \in P \times L : p \in l \} | = e ( S , T)
\]
where $e(S,T)$ is the number of edges in $ \mathcal{G}( \Pi )$ with one endpoint in $S$ and the other in $T$.
By Lemma \ref{beml},
\[
| \{ (p , l ) \in P \times L : p \in l \} | \leq \frac{ |S| |T| }{q} + q^{1/2}\sqrt{|S||T|}
\]
which proves Theorem \ref{main th2}.
\end{proof}

\bigskip

\begin{proof}[Proof of Theorem \ref{main th} and Corollary \ref{main cor}]
Let $A \subset Q^*$.  Let $S = (A + A ) \times (A \cdot A)$.  We view $S$ as a subset of $\mathcal{P}$.  Let $s(a,b) = \{ (x , y) \in Q^2 : y = b \cdot  x- b \cdot a \}$ and
\[
L = \{ s (a,b) : a,b \in A \}.
\]
By Lemma \ref{l3}, $|L| = |A|^2$ and there are at least $|A|^3$ incidences between $S$ and $L$.
Let $T = \{ [ - b , -b \cdot a ] : a,b \in A \}$ so $T$ is a subset of $\mathcal{L}$.  By Lemma \ref{beml},
\[
e(S,T) \leq \frac{ |S| |T| }{q} + q^{1/2}\sqrt{|S||T|}.
\]
We have $|L| = |T| = |A|^2$.  If $m = |A+A|$ and $n = |A \cdot A|$, then
\[
e(S,T) \leq \frac{ mn |A|^2 }{q} + q^{1/2}|A|\sqrt{mn}.
\]
Next we find a lower bound on $e(S,T)$.  By construction, an incidence between $S$ and $L$ corresponds to an edge between $S$ and $T$ in $ \mathcal{G}( \Pi )$.  To see this, note that $(x,y) \in S$ is incident to $s(a,b) \in L$ if and only if $y = b \cdot x - b \cdot a$.
This is equivalent to the equation $- b \cdot x + y = - b \cdot a$ which holds if and only if $(x,y)$ is adjacent to $[-b , -b \cdot a]$
in $ \mathcal{G}( \Pi)$.  Thus,
\begin{equation}\label{eq:eq:1}
|A|^3 \leq e(S,T) \leq  \frac{ mn |A|^2 }{q} + q^{1/2}|A|\sqrt{mn}.
\end{equation}

To prove Corollary \ref{main cor}, observe that from (\ref{eq:eq:1}), we have
\[
| A + A | |A \cdot A | \geq \textup{min} \left\{ c q |A| , \frac{ c |A|^4 }{q} \right\}
\]
where $c$ is any real number with $c + c^{1/2} < 1$.  If $x = \max \{ |A+A| , |A \cdot A | \}$, then
$ x \geq \textup{min} \{ ( cq |A| )^{1/2} , \frac{ c^{1/2} |A|^2 }{q^{1/2} } \}$ and Corollary \ref{main cor} follows from this inequality.
\end{proof}

\bigskip

\begin{proof}[Proof of Corollary \ref{v2:cor 5}]
Let $A \subset Q$, $P = A \times ( A \cdot (A + A ) )$,
\[
l( b,c) = \{ (x,y) \in Q^2 : y = b \cdot ( x+c) \},
\]
and $L = \{ l(b,c) : b,c \in A \}$.  Then $|P| = |A| | A \cdot (A + A) |$, $|L| = |A|^2$, and
$L$ is a set of lines in $Q^2$.  Let $z = |A \cdot (A + A)|$.  Observe that each $l(b,c) \in L$ contains at least $|A|$ points from $P$.  By Theorem
\ref{main th2},
\[
|A|^3 \leq \frac{ |P| |L| }{q} + q^{1/2} \sqrt{ |P| |L| } =
\frac{ |A|^3 z}{q} + q^{1/2} |A|^{3/2} z^{1/2}.
\]
This implies that $q |A|^{3/2} \leq |A|^{3/2} z  + q^{3/2} \sqrt{z}$.
Therefore,
\[\sqrt{z}\ge \frac{-q^{3/2}+\sqrt{q^3+4|A|^{3}q}}{2|A|^{3/2}}=\frac{4|A|^3q}{2|A|^{3/2}(q^{3/2}+\sqrt{q^3+4|A|^3q})},\]
which implies that
\[
| A \cdot (A + A) | \geq c \min \left\{ q , \frac{ |A|^3 }{q} \right\}.
\]
We note that if $|A| \gg q^{2/3}$ then we can take $c = 1+o(1)$.
\end{proof}

\bigskip

\begin{proof}[Proof of Theorem \ref{main th3} and Corollary \ref{main cor3}]
Let $A , B , C, D \subset Q$.  Consider the sets $P = \{ ( d , -a ) : d \in D, a \in A \}$
and $L = \{ [ c , b + \gamma ] : c \in C , b \in B \}$.  An edge between $P$ and $L$ in $ \mathcal{G}( \Pi )$ corresponds to a solution to
$c \cdot d + (-a) = b + \gamma$ with $c \in C$, $d \in D$, $a \in A$, and $b \in B$.  Therefore,
$ e( P , L)$ is precisely the number of solutions to $a + b + \gamma = c \cdot d$ with
$(a,b,c,d) \in A \times B \times C \times D$.
Observe that
$|P | = |D| |A|$ and $|L| = |C| |B|$.  By Lemma
\ref{beml},
\[
\left| N_{ \gamma} (A,B,C,D)   - \frac{ ( q + 1) |A| |B| |C| |D| }{q^2 + q + 1} \right| \leq q^{1/2} \sqrt{ |A| |B| |C| |D| }.
\]

To obtain Corollary \ref{main cor3}, apply Theorem \ref{main th3} with $A$, $B$, $C$, and $-D$.  For any
$- \gamma \in Q$, the number of $(a,b,c,-d) \in A \times B \times C \times (-D)$ with
$a+b - \gamma = c \cdot (-d)$ is at least
\begin{equation}\label{v2:eq1}
\frac{ (q+1) |A| |B| |C| |-D| }{q^2 + q + 1} - q^{1/2} \sqrt{ |A| |B| |C| |-D| }.
\end{equation}
When $|A| |B| |C| |D| > q^3$, (\ref{v2:eq1}) is positive and so we have a solution
to $a+b - \gamma = c \cdot (-d)$.  Since this equation is equivalent to
$a + b + c \cdot d = \gamma$ and $\gamma$ was arbitrary, we get
\[
Q  = A + B + C \cdot D.
\]
\end{proof}

\section{Proof of Theorems  \ref{v2:theorem 4}, \ref{v2:theorem 1}, \ref{v2:theorem 2}, and \ref{v2:theorem 3}}

Let $Q$ be a finite quasifield with $q$ elements and let $K$ be the kernel of $Q$.  Let $d \geq 1$ be an integer and $\gamma \in Q$.
The \emph{product graph}, denoted
$\mathcal{DP}_Q(d, \gamma)$, is the bipartite graph with parts $X$ and $Y$ where $X$ and $Y$ are disjoint copies
of $Q^{d+1}$.
The vertex $(x_1, \dots , x_{d+1})_X \in X$ is adjacent to $(y_1, \dots , y_{d+1})_Y \in Y$ if and only if
\begin{equation}\label{v2:eq1 for theorem}
x_{d+1} + \gamma  = x_1 \cdot y_1 + x_2 \cdot y_2 + \dots + x_d \cdot y_d +  y_{d+1}.
\end{equation}

\begin{lemma}\label{v2:l1 for theorem}
The graph $\mathcal{DP}_Q (d , \gamma )$ is $q^d$-regular.
\end{lemma}
\begin{proof}
Fix a vertex $(x_1, \dots , x_{d+1})_X \in X$.  We can choose $y_1, \dots , y_d$ arbitrarily and then (\ref{v2:eq1 for theorem}) gives a unique solution for $y_{d+1}$.  Therefore, $(x_1, \dots , x_{d+1})_X$ has degree $q^d$.  A similar argument
shows that every vertex in $Y$ has degree $q^d$.
\end{proof}

\begin{lemma}\label{v2:l2 for theorem}
If $\lambda_1 \geq \lambda_2 \geq \dots \geq \lambda_n$ are the eigenvalues of
$\mathcal{DP}_Q (d , \gamma )$, then $| \lambda | \leq q^{d/2}$ where $\lambda = \max_{i \neq 1,n} | \lambda_i |$.
\end{lemma}
\begin{proof}
Let $M$ be the adjacency matrix
of $\mathcal{DP}_Q ( d , \gamma)$.  Assume that the first $q^{d+1}$ rows/columns of $M$ correspond to the vertices of $X$.  We can write
\[
M = \begin{pmatrix} 0 & N \\ N^T & 0 \end{pmatrix}
\]
where $N$ is the $q^{d+1} \times q^{d+1}$ matrix whose $(x_1 , \dots , x_{d+1})_X \times (y_1 , \dots , y_{d+1})_Y$-entry is
1 if (\ref{v2:eq1 for theorem}) holds, and is 0 otherwise.  Let $J$ be the $q^{d+1} \times q^{d+1}$ matrix of all 1's and
let
\[
P = \begin{pmatrix} 0 & J \\ J & 0 \end{pmatrix}.
\]
We claim that
\begin{equation}\label{v2:l2 eq}
M^3 = q^d M + q^d (q^{d-1} - 1)P.
\end{equation}
The $(x,y)$-entry of $M^3$ is the number of walks of length 3 from
$x = (x_1 , \dots , x_{d+1})_X$ to $y= (y_1, \dots , y_{d+1})_Y$.  Suppose that $x y' x' y$ is such a walk where
$y' = (y_1 ' , \dots ,  y_{d+1}' )_Y$ and $x' = (x_1 ' , \dots , x_{d+1} ' )_X$.  By Lemma \ref{v2:l1 for theorem}, there are $q^d$
vertices $x' \in X$ such that $x'$ is adjacent to $y$.  In order for $x y' x' y$ to be a walk of length 3, $y'$ must be adjacent to
both $x$ and $x'$ so we need
\begin{equation}\label{v2:l2 eq2}
x_{d+1} = x_1 \cdot y_1 ' + \dots +  x_d \cdot y_d ' + y_{d+1} '
\end{equation}
and
\begin{equation}\label{v2:l2 eq3}
x_{d+1} ' = x_1 ' \cdot y_1 ' +  \dots + x_d ' \cdot y_d ' + y_{d+1} '.
\end{equation}
We want to count the number of $y'$ that satisfy both (\ref{v2:l2 eq2}) and (\ref{v2:l2 eq3}).  We consider two cases.

\medskip
\noindent
\textit{Case 1}: $x$ is not adjacent to $y$.

If $x_i = x_i ' $ for $1 \leq i \leq d$, then (\ref{v2:l2 eq2}) and
(\ref{v2:l2 eq3}) imply that $x_{d+1} = x_{d+1} '$.  This implies $x = x'$ and so $x$ is adjacent to $y$ but this contradicts our assumption that $x$ is not adjacent to $y$.  Therefore, $x_i \neq x_i '$ for some $1 \leq i \leq d$.   Without loss of generality, assume that $x_1 \neq x_1 '$.  Subtracting (\ref{v2:l2 eq3}) from (\ref{v2:l2 eq2}) gives
\begin{equation}\label{v2:l2 eq4}
x_{d+1} - x_{d+1} ' = x_1 \cdot y_1 ' + \dots + x_d \cdot y_d ' - x_1 ' \cdot y_1 ' - \dots - x_d ' \cdot y_d '.
\end{equation}
Choose $y_2 ' , \dots , y_d ' \in Q$.  Since $Q$ is a quasifield and $x_1 - x_1' \neq 0$, there is a unique solution for $y_1 '$ in
(\ref{v2:l2 eq4}).  Equation (\ref{v2:l2 eq2}) then gives a unique solution for $y_{d+1}'$ and so there are
$q^{d-1}$ choices for $y' = (y_1 ' , \dots , y_{d+1} ' )_Y$ for which both (\ref{v2:l2 eq2}) and (\ref{v2:l2 eq3}) hold.  In this case, the number of walks of length 3 from $x$ to $y$ is $(q^d - 1) q^{d-1}$ since $x'$ may be chosen in $q^d - 1$ ways as we require
$(x_1 '  , x_2 ' , \dots , x_d '  ) \neq (x_1 , x_2 , \dots , x_d)$ otherwise $x = x'$.

\medskip
\noindent
\textit{Case 2}: $x$ is adjacent to $y$.

The same counting as in Case 1 shows that there are
$(q^d - 1)q^{d-1}$ paths $xy' x' y$ with $x \neq x'$.  By Lemma \ref{v2:l1 for theorem}, there are $q^d$ paths of the form $xy' x y$ since the degree of $x$ is $q^d$.

\medskip

From the two cases, we deduce that
\[
M^3 = q^d M + q^{d-1} (q^d - 1)P.
\]
Let $\lambda_j$ be an eigenvalue of $M$ with $j \neq 1$ and $j \neq n$.  Let $v_j$ be an eigenvector for $\lambda_j$.  Since $v_j$ is orthogonal to $\chi_X + \chi_Y$ and $\chi_X - \chi_Y$, we have $Pv_j = 0$ and so
\[
M^3 v_j = q^d M v_j.
\]
This gives $\lambda_j^3 = q^d \lambda_j$ so $| \lambda_j | \leq q^{d/2}$.
\end{proof}

\bigskip

\begin{proof}[Proof of Theorem \ref{v2:theorem 4}]
Let $A , B , C \subset Q$ where $Q$ is a finite quasifield with $q$ elements.  Given $\gamma \in Q$, let
\[
Z_{ \gamma } = \{ ( a , b ,c ) \in A \times B \times C : a + b \cdot c = \gamma \}.
\]
We have $\sum_{ \gamma } |Z_{ \gamma} | = |A| |B| |C|$ so by the Cauchy-Schwarz inequality,
\begin{equation}\label{v2:eq 1 for th4}
|A|^2 |B|^2 |C|^2 = \left( \sum_{ \gamma } | Z_{ \gamma} | \right)^2 \leq
|A + B \cdot C | \sum_{ \gamma  \in Q} |Z_{ \gamma} |^2.
\end{equation}
Let $x = \sum_{ \gamma} |Z_{ \gamma }|^2$.  By (\ref{v2:eq 1 for th4}),
\begin{equation}\label{v2:eq 1 for th5}
| A + B \cdot C| \geq \frac{ |A|^2 |B|^2 |C|^2 }{x}.
\end{equation}
The integer $x$ is the number of ordered triples $(a,b,c)$, $(a' , b' , c')$ in $A \times B \times C$ such that
$a+ b \cdot c = a' + b' \cdot c'$.  This equation can be rewritten as
\[
a =  b \cdot (- c) + b' \cdot c'  + a'.
\]
This is equivalent to the statement that $(b,b',a)_X$ is adjacent to $(-c,c',a')_Y$ in the graph $\mathcal{DP}_Q ( 2 , 0)$.
Thus, $x$ is the number of edges between the sets
\[
S = \{ (b,b',a)_X : a \in A , b ,b' \in B \}
\]
and
\[
T = \{ ( -c , c' , a' )_Y : a' \in A , c , c ' \in C \}
\]
in $\mathcal{DP}_Q (2, 0)$.  By Lemma \ref{beml} and Lemma \ref{v2:l2 for theorem},
\[
x = e(S,T) \leq \frac{ |S| |T| }{q} + q^{1/2} \sqrt{ |S| |T| }.
\]
This inequality together with (\ref{v2:eq 1 for th5}) gives
\[
\frac{ |A|^2 |B|^2 |C|^2}{ |A + B \cdot C| } \leq x \leq \frac{ |A|^2 |B|^2 |C|^2 }{q}  + q |A| |B| |C|
\]
from which we deduce
\[
|A + B \cdot C | \geq q - \frac{q^3}{|A||B||C| + q^2}
\]
\end{proof}

\bigskip

\begin{proof}[Proof of Theorem \ref{v2:theorem 1}]
Let $A \subset Q$,
$S =  A \times  \cdots \times A \times (-A) \subset Q^{d+1}$, and
 $T =  A \times \cdots \times A \times A \subset Q^{d+1}$.  View $S$ as a subset $X$ and $T$ as a subset of
$Y$ in the graph $\mathcal{DP}_Q ( \gamma , d)$.
By Lemmas \ref{beml} and \ref{v2:l2 for theorem},
\[
\left| e(S , T) - \frac{q^d |S| |T| }{q^{d+1} } \right| \leq 2 q^{d/2} \sqrt{ |S| |T| }.
\]
An edge between $S$ and $T$ corresponds to a solution to
\[
- a_{d+1} + \gamma = a_1 \cdot a_1 ' + \dots + a_d \cdot a_d ' + a_{d+1}'
\]
with $a_i , a_i ' \in A$.  If $|A| > q^{ \frac{d+2}{2d+2} }$, then $e(S,T) > 0$.  Since $\gamma$ is an arbitrary element of $Q$, we get
\[
Q  = A +A + \underbrace{A \cdot A + \dots + A \cdot A}_{d~\textup{terms}}
\]
which completes the proof of Theorem \ref{v2:theorem 1}.
\end{proof}

\bigskip

\begin{proof}[Proof of Theorem \ref{v2:theorem 2}]
We will work in the graph $\mathcal{DP}_Q ( 2 , 0)$.
Let $\gamma \in Q^*$ and $A,B,C,D \subset Q$.  For each pair $(b,d ) \in B \times D$, define
\[
L_{ \gamma} ( b , d) = \{ ( b \cdot \lambda , d \cdot \lambda , - \gamma \cdot \lambda )_Y : \lambda \in K^* \}.
\]

\medskip
\noindent
\textit{Claim 1}: If $(a,c) \in A \times C$ and $a \cdot b + c \cdot d = \gamma$, then $(a,c,0)_X$ is adjacent to every vertex in
$L_{ \gamma } ( b, d)$.

\smallskip
\noindent
\textit{Proof.} Assume $(a,c)\in A \times C$ satisfies $a \cdot b  + c \cdot d = \gamma$.  If
$\lambda \in K^*$, then
\[
a \cdot ( b \cdot \lambda ) + c \cdot ( d \cdot \lambda ) = ( a \cdot b) \cdot \lambda + ( c \cdot d ) \cdot \lambda
= (a \cdot b + c \cdot d ) \cdot \lambda = \gamma \cdot \lambda.
\]
Therefore, $0 = a \cdot ( b \cdot \lambda) + c \cdot ( d \cdot \lambda) - \gamma \cdot \lambda$ which shows that
$(a,c,0)_X$ is adjacent to $( b \cdot \lambda , d \cdot \lambda , - \gamma \cdot \lambda )_Y$.

\bigskip
\noindent
\textit{Claim 2}: If $(b_1 , d_1 ) \neq (b_2 , d_2)$, then $L_{ \gamma } ( b_1 , d_1 ) \cap L_{ \gamma } (b_2 , d_2 ) = \emptyset$.

\smallskip
\noindent
\textit{Proof.}  Suppose that $L_{ \gamma } (b_1 , d_1 ) \cap L_{ \gamma } (b_2 , d_2 ) \neq \emptyset$.  There are elements $\lambda , \beta \in K^*$ such that
\[
( b_1 \cdot \lambda , d_1 \cdot \lambda , - \gamma \cdot \lambda )_Y
= (b_2 \cdot \beta , d_2 \cdot \beta ,  - \gamma \cdot \beta )_Y.
\]
This implies
\[
b_1 \cdot \lambda = b_2 \cdot \beta ,~~ d_1 \cdot \lambda  = d_2 \cdot \beta , ~~ \mbox{and}~~ \gamma \cdot \lambda = \gamma \cdot \beta.
\]
Since $\gamma \cdot \lambda = \gamma \cdot \beta$, we have $\gamma \cdot ( \lambda  - \beta )  = 0$.  As
$\gamma \neq 0$, we must have $\lambda = \beta$ so $b_1 \cdot \lambda = b_2 \cdot \beta = b_2 \cdot \lambda$.
Using Lemma \ref{v2:l0},
\[
0 = b_1 \cdot \lambda - ( b_2 \cdot \lambda) = b_1 \cdot \lambda + ( - b_2 ) \cdot \lambda
= (b_1 - b_2 ) \cdot \lambda.
\]
Since $\lambda \neq 0$, we have $b_1 = b_2$.  A similar argument shows that $d_1 = d_2$.

\bigskip

Let $S = \{ (a,c,0)_X : a \in A , c \in C \}$ and
\[
T =  \bigcup_{ (b,d) \in B \times D} L_{ \gamma} ( b,d).
\]
The number of edges between $S$ and $T$ in $\mathcal{DP}_Q(2 , 0)$ is $N_{\gamma } ( |K| -1)$ where $N_{ \gamma}$ is the number of 4-tuples $(a,b,c,d) \in A \times B \times C \times D$ such that
$a \cdot b + c \cdot d = \gamma$.  Furthermore
$|S| = |A| |C|$ and $|T| = |B| |D| ( |K| - 1)$ by Claim 2.  By Lemmas \ref{beml} and \ref{v2:l2 for theorem},
\begin{equation}\label{v2:eq100}
\left| N_{ \gamma } ( |K| - 1) - \frac{ |S| |T| }{q} \right| \leq q \sqrt{ |S| |T| }.
\end{equation}
This equation is equivalent to
\[
\left| N_{ \gamma}  - \frac{ |A| |B| |C| |D| }{q} \right| \leq q \left( \frac{ |A| |B| |C| |D| }{ |K| - 1 } \right)^{1/2}
\]
which completes the proof of Theorem \ref{v2:theorem 2}.
\end{proof}

\bigskip

The proof of Theorem \ref{v2:theorem 3} is similar to the proof of Theorem \ref{v2:theorem 2}.  Instead of working with the graph
$\mathcal{DP}_Q(2,0)$, one works with the graph $\mathcal{DP}_{Q}(d,0)$.
One counts edges between the sets
\[
S = \{ (a_1 ' , \dots , a_d ' , 0 )_X : a_i ' \in A \}
\]
and
\[
T = \bigcup_{ (a_1 , \dots , a_d) \in A^d } L_{ \gamma } ( a_1 , \dots , a_d)
\]
where $L_{ \gamma }(a_1 , \dots, a_d) = \{ (a_1 \cdot \lambda , \dots , a_d \cdot \lambda , - \gamma \cdot \lambda )_Y : \lambda \in K^* \}$.  The remaining details are left to the reader.

redpajama_set_name: RedPajamaArXiv


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text:
Tag: Paul Dame
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Vermont GOP set to nominate candidates for auditor, attorney general; still searching for treasurer
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Windham County Senate field rises to 6 candidates with party switch
@finitecapacity and Facebook/Todd Hemingway
 " data-medium-file="https://vtdigger.org/wp-content/uploads/2022/07/mark-coester-colchester-parade-300x228.jpg" data-large-file="https://vtdigger.org/wp-content/uploads/2022/07/mark-coester-colchester-parade-610x464.jpg" />
The unofficial top Republican vote-getter in last week's primary, disavowed by Vermont GOP leaders for displaying fascist and alt-right symbols, will run as an independent.
Vermont GOP declines to back US House nominee Liam Madden
The state party says it will withhold support for Liam Madden's Republican congressional bid because he will not commit to caucusing with Republicans in Washington.
Republican field in Windham County Senate race remains unclear
Mark Coester, disavowed by Vermont GOP leaders for displaying fascist and alt-right symbols, may drop his bid for one of two open seats and instead run as an independent for U.S. Senate.
In the Republican lieutenant governor primary, one candidate sees an existential crisis. The other doesn't 'buy into that.'
Statehouse veteran and moderate Republican Joe Benning is up against Gregory Thayer, a former local GOP chair who participated in the Jan. 6, 2021, riot in Washington, D.C. in Vermont's Republican primary for lieutenant governor.
Candidate draws condemnation for displaying controversial symbols at Colchester 4th of July celebration
By Lia Chien
Town officials have disavowed what appeared to be a fascist flag and an alt-right symbol brandished by Mark Coester at the annual parade.
Winooski's noncitizen voting expected to survive lawsuit
By Dominic Minadeo
Jun 26 2022 June 27, 2022
The national and state arms of the Republican Party filed suit against Winooski and Montpelier last fall, arguing that charter changes in the two cities affording noncitizens the right to vote in local elections were unconstitutional.
The paperwork is in. Here are (most of) the legislative races to watch.
Jun 5 2022 June 6, 2022
Battleground districts in Franklin County and across the state could determine whether Democrats hold the power to override the governor's vetoes in the next legislative session.
Republican Kevin Divney withdraws from state treasurer's race days after DUI charge surfaces
Perennial candidate H. Brooke Paige is now the only Republican primary contender in Vermont's races for treasurer, auditor, secretary of state and attorney general.
In election season with historic turnover, some statewide races will see little competition
May 26 2022 June 24, 2022
Thursday was the deadline for Vermont political hopefuls to file their primary candidacy petitions with the Secretary of State's Office.

redpajama_set_name: RedPajamaCommonCrawl


text:
Vol. 6, No. 1 / Screens / The Courtyard Commission: Dispersion
Curatorial Statement – The Courtyard Commission: Dispersion
By Scheherazaad Cooper
Photo credit: Jevan Crittenden.
The Courtyard Commission emerged from a desire to re-imagine and re-envision the potential of the beautiful courtyard space at the Ismaili Centre, Burnaby. What happens when a group of innovative artists are brought into an inspiring space? What will they see, and what will it inspire them to create? How will a courtyard whose design draws from the traditions of Islamic gardens— built initially for an immigrant community developing roots in Canada—speak to others who may not have direct access to the significance the space holds for a particular community?
More than just a performance, the first Courtyard Commission was very much about engagement. Mulitmedia artists Nancy Lee, Laine Butler, and Kiran Bhumber created and worked within the space over the course of a summer residency. Working also with Choreographer Ralph Escamillian and dancers Sophia Wolfe and Alex Tam, the team learned new things about their own practices; they developed bespoke software and acquired new skills in order to actualize the vision that emerged from their residency and, thus, the remarkable piece Dispersion came to be.
Dispersion, the second performance in The Courtyard Commission, was performed in the Fall of 2018. It contributes to an emerging canon of performance in the Lower Mainland, where the building and architecture is an intrinsic part of the performance and the spectator is encouraged to move freely around the space and to take in the work from multiple viewpoints.
During the performances, I was approached by a number of other local artists who were in awe of the space and were beginning to interpret it in creative ways, themselves. Furthermore, spectators had an opportunity to experience the affect the courtyard had on these artists, we saw it through their eyes—and this is a deeply significant element.
My hope is that this project has opened up, at a number of levels, opportunities for artists to re-envision and re-interpret established and/or socially significant architectural spaces to expand the ways they are inhabited, experienced, and animated.
Scheherazaad Cooper is an accomplished Canadian performer and choreographer, drawing from nearly thirty years as an Odissi classical Indian dancer, as well as a background in theatre, in her contemporary work.
Rungh Redux Winner 2022 Award of Merit Innovative Practice
Britannia Art Gallery
Bookhug Press
Plantation Memories
Alternator Centre


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Летючий човен — тип літака-гідроплана з фіксованим крилом, нижня частина корпусу якого виготовлена у вигляді човна, пристосованого для швидкого пересування по поверхні води. Летючий човен, як правило, не має відповідного посадкового устаткування, яке б дозволяло йому приземлюватися на суходолі. Він відрізняється від поплавкового гідролітака тим, що використовує спеціальний фюзеляж, який може плавати, забезпечуючи плавучість літака. Летючі човни можуть стабілізуватися на водній поверхні за допомогою підкрильних поплавків або крилоподібних виступів (т. зв. спонсонів) у нижній частині фюзеляжу. Протягом першої половини XX століття летючі човни були одними з найбільших літаків, за розмірами їх перевищували лише бомбардувальники, розроблені під час Другої світової війни. Їхньою перевагою була спроможність використовувати воду замість коштовних наземних злітно-посадкових смуг, що робило їх основою для міжнародних авіакомпаній у міжвоєнний період. Вони також зазвичай використовувалися для морського патрулювання та рятування на морі.

Конструкція
Створення летючих човнів було викликане необхідністю збільшення габаритних розмірів гідролітаків з метою розміщення пасажирів і обладнання.

Летючі човни поділяються на одночовнові, двочовнові (типу «катамаран») і інтегральну схеми літальних апаратів (остання найбільш доцільна для важких багатоцільових океанських гідролітаків). Крім високих аеродінамічних та злітно-посадочних характеристик звичайного літака, летючий човен забезпечує основні морехідні якості (плавучість, поздовжню і поперечну остійність, глісування, керованість, непотоплюваність тощо), тобто можливість експлуатації в акваторіях за певних гідрометеорологічних умов (швидкість і напрям вітру, форма, висота і довжина хвиль води та інше).

Для досягнення оптимальніших характеристик летючим човнам надають плавно обтічну, загострену до носа і корми форму, що відрізняється формою днища у вигляді кіля, з різко вираженим хвилерізом у носовій частині. На днище човна розташовуються поперечні редани (уступи на днище летючого човна для зриву водяного потоку або струменів), один — поблизу центру інерції, другий — у кормовій частині човна. З метою підвищення мореплавства на носовій частині днища човна іноді розміщують поздовжні редани.

Внаслідок високого розташування центру інерції одночовнова схема неостойчива в поперечній площині, тому вона забезпечується двома підкрильних поплавками чи зябрами (пластини трапецієподібної форми, які кріпляться до бортів летючого човна в середній частині, забезпечуючи в такий спосіб поперечну остійність). Для додання непотоплюваності внутрішній об'єм летючого човна розбивають водонепроникними перегородками на герметичні відсіки, що дозволяють човнові зберігати плавучість і остійність при пошкодженні конструкції і затопленні окремих відсіків.

Див. також
 Екраноплан
 Летючий підводний човен
 Літак-амфібія
 Патрульний літак
 Морська розвідка

Посилання

 ЛЕТА́ЮЩАЯ ЛО́ДКА
 Гидросамолеты, гидропланы и самолеты-амфибии: что собой представляют

Відео
 The Flying Boats full documentary

Класифікація літаків
Гідролітаки

redpajama_set_name: RedPajamaWikipedia


I did occasionally sit down to sketch and paint while in Spain - while listening to music, sipping coffee, watching Flamingo dancers, or drinking a glass of wine. The sites weren't pre-selected and all but once, my husband was with me as we talked and enjoyed our vacation.
I walked to Plaza Santa Ana one sunny afternoon while my husband went to the gym. There was a place to sit in a central area among outdoor cafes, each with their own musicians that alternated playing. The Teatro Espanol was at one end of the Plaza and I sat there enjoying my surroundings while drawing and painting.
My husband and I took a long walk our last day in Madrid, and stopped for a mid-morning cup of coffee. I was facing several apartments around the Plaza San Andreas, and I loved the combination of colors and the sun vs shade.
Seville: We had dinner in our small hotel dining room the day we arrived, and less than a dozen of us watched a pair of Flamenco dancers, a guitarist, and a singer. The flamenco dancers moved too quickly to sketch, so I drew one of the repeated poses of the guitarist.
We spent one entire day looking at Antoni Gaudi buildings: Sagrada Familia, Guadi's home and museum at Parc Guell, and Casa Batllo several blocks from our hotel. After walking around the exterior and interior of Sagrada Familia, I set up front, in the nave, and sketched the ceiling above the altar while listening to someone, somewhere sing Ave Maria.
We stopped for a glass of wine in a gorgeous plaza, after visiting the Picasso Museum. While there, a well-known elderly Flamenco Dancer stopped right in front of us, so he and the students made it into my painting.
Before we left for Spain, I put 2 X 2" squares on several sketchbook pages, so I could add one image that stood out in my mind each day. Here are the 3 grid pages that are now completed.
I also prepared pages to record activities for each day, which always is appreciated when I'm trying to remembers restaurants and other places.
We just returned from 12 days in Madrid, Seville, and Barcelona. It was hard to find time to sit down to do a quick sketch as we traveled, so some of my time I carried an open sketchbook and sketched our walking tour guides, and an interesting roofline or something to remind me of the places. Each tour (except Santa Cruz in Seville) is about 3 hours and 2-3 miles. The Santa Cruz tour was as long as the others, but that area which is very old is very small with winding alleys. I have 5 of these journal pages, one with a watercolor wash, and 4 without. Since this is "Inktober Month" I may leave them unpainted.
Walking Tour of Madrid - from Plaza Mayor, through old Madrid and then up to the Palacio Real, Cathedral Almudena, and Opera.
Seville Walking Tour of Barrio de Santa Cruz - the tiny old section that was occupied by the Muslims and Jews from at least the 8th C until 1492 when Ferdinand and Isabella expelled all non-Catholics from Spain.
Walking Tour - Highlights of Seville: This tour began at the Seville Cathedral, then extended to the Real Alcazar, the Archives of the Indies, the Harbor, the Bullring, and the Spanish Pavillion.
Granada: We took a day trip from Seville to Granada to see The Alhambra - lots of hours on a bus, but worth it.
Barcelona - We loved the city, but had a very disappointing tour guide. We've done these walking tours with many companies in many cities, and this was the only one that we left before the end, partly because it started to rain!
We were traveling with another couple, friends for 45+ years. Only my husband was in Spain previously and then only in Madrid for a few days for a meeting. So glad we finally saw these 3 big cities, and hopefully we will return to Barcelona to explore this amazing city in the future.
Several weeks ago I went to a Fashion Illustration session organized by Simon Levinson at the National Arts Club on Gramercy Park. Simon is the leader of the Meetup Group called Drawing New York. I had no idea what to expect, but it was great fun. Figure Al Fresco prepared me for drawing clothed figures, but not the exaggerated proportions of fashion illustrations. It was fun to see the stylized drawings of some of the experts.
Here are the main models, certainly part of a time warp. These photos came from the event website.
The first pose was 10 minutes, the next 4 were 15 minutes, and the last one was 10 minutes. I sketched on paper that was too large to scan, so these were photographs.
A beautiful blond woman, that was dressed looking like a man in the clothes she wore. But the shoes are gender specific.


text:
Oct. 7, 2002 / 5:11 PM
Scott's World -- UPI Arts & Entertainment
VERNON SCOTT, United Press International
HOLLYWOOD, Oct. 7 (UPI) -- It's difficult to believe a jailed star should suddenly find episodes of his TV series being sold on the eve of his trial for murdering his wife.
The actor is Robert Blake, 69, charged with shooting Bonnie Lee Bakley, 44, to death in the family car in May 2001.
Episodes of Blake's show, "Baretta" -- in which he played a street-smart cop (1975-1978) -- can now be purchased in stores.
Baretta was a caricature of the actor, a sloppily dressed undercover tough guy specializing in disguises, living in a rundown hotel with a pet cockatoo. His wardrobe: grungy T-shirt and denims.
Is it coincidence that Universal Studios chose to include "Baretta" in its DVD releases of the studio's most popular TV programs?
Other shows in the series include "Battlestar Galactica," "Quantum Leap," "Dragnet," "Magnum P.I." and "The Rockford Files." None of the stars involved is likely to be as interesting as Blake, who has been behind bars since last March awaiting trial.
He has already pleaded innocent, but Los Angeles Police Department media releases read as if he already was condemned by a judge and jury -- which is not the case.
"Baretta" is prominent in Universal's press releases with two DVD's: "Baretta Season One" (the first 13 shows) and "The Best of Baretta."
Was Blake's name in headlines a factor at this time? Odds are good "Baretta" will top the list in sales, even with a price of $39.98.
The other stars in this crime/detective category are Tom Selleck, Peter Falk, James Garner, Scott Bakula -- none of whom is involved in a sensational murder case.
Blake increasingly becomes hot copy as his trial approaches. Try to imagine the outcry and indignation had some studio released the movies or TV appearances of O.J. Simpson on the eve of his trial for the murder of his former wife, Nicole.
Universal Home Studio Video might not have considered the possibility of exploiting Blake in his private nightmare, but if it is coincidence, the release date this week appears to be propitious.
In its press release, Universal lauds Blake's show:
"Robert Blake found his true TV persona, a rough-and-tumble, unconventional cop, often at odds with the police brass.
"Featuring hip soundtrack and music from Sammy Davis, Jr., and Simon & Garfunkle and ... guest stars Gary Busey, M. Emmett Walsh, Scott Glenn, Tommy Lee Jones, Wings Hauser, Jennifer Jason Leigh and Dennis Quaid ... Baretta stands today as a program against which all other cool cops shows must be measured."
High praise indeed considering the studio's relations with Blake from beginning to end were contentious; Blake bitterly protested the real or imagined poor treatment his show received from studio brass.
The studio suffered many truculent Blake broadsides as the show became increasingly popular; the higher the ratings the more difficult and demanding Blake became.
Finally the show was canceled after many an acrimonious allegation and counter-charge.
It was David vs. Goliath, and this time little David took a heavy loss. Blake was fired by the powerful MCA conglomerate.
Without question, "Baretta" was the high-water mark of Blake's career, although he has appeared in 131 movies and television shows, going back to his days as an appealing little waif in the "Our Gang" films.
His biggest movie hits were "In Cold Blood," "Electra Glide in Blue" and "Tell Them Willie Boy Is Here."
Always a moody loner with a chip on his shoulder, Blake was a perpetual outsider in Hollywood, shunning glitzy parties and pompous industry events. He carried his impoverished childhood around heavily on his muscular shoulders, taking frequent forlorn midnight walks alone in the dry Los Angeles River bed near his Studio City home.
Blake, as he survives the long hours of solitude in his barred Los Angeles jail cell, must reflect with misgivings the irony of Baretta's maxim, "Don't do the crime if you can't do the time."
It was one many adages Blake contributed to scripts during the "Baretta" series, one he often repeated in conversations about crime. Now Blake finds himself doing time in the clink since last April, awaiting his trial for murder. Currently he is scheduled for a bail hearing Nov. 13.
His attorneys have produced evidence that Blake's wife, Bakley, corresponded with more than 140 prison inmates, including serial child molesters and murderers nationwide, in which she used some 130 aliases. The implication: any one of the criminals might have a motive for killing her.
Among Bakley's correspondents was Christian Brando, son of actor Marlon Brando, who dated Bakley while she was Blake's companion.
Brando once told Bakley he had impregnated her, but DNA tests proved Blake was the father of little Rose, now 2, who lives with Blake's grown daughter in a secluded L.A. suburb.
The new DVD's should sell briskly.
Sammy Davis


text:
Retail Dining Developments
Lucky's Market named as new anchor for Apopka retail center redevelopment
Marketing site plan for the Wekiva Riverwalk shopping center in Apopka. (Woolbright)
New development, construction and marketing plans affirm Lucky's Market is pursuing its eighth Greater Orlando store in the Wekiva Riverwalk shopping center in Apopka later this year.
The Apopka City Council will vote next Wednesday on the final Development Plan for the project, which involves demolition and then redevelopment of the northeast corner of the shopping center at Wekiva Springs Road and East Semoran Boulevard (S.R. 436). On Tuesday, the city's Planning and Zoning Commission recommended approval of the plan.
The applicant is Boca-Raton-based Woolbright Development, which owns Wekiva Riverwalk and will handle redevelopment and leasing. Woolbright and Lucky's Market declined to comment on Thursday.
Lucky's Market signs for 8th area store with Unicorp's Griffin Farm in Lake Mary
Your update on construction progress at this 33-acre mixed-use development, and the flurry of new site activity in Greater Orlando by this specialty grocer.
By Bob Moser
Feb 09, 2018 | 8:47 AM
GrowthSpotter first reported in late January on the Woolbright-Lucky's pending contract.
Apopka city officials this week, and Woolbright marketing materials, confirm Lucky's Market as one of the new anchor tenants in Wekiva Riverwalk.
Plans Woolbright submitted to the St. Johns River Water Management District this month show the company wants to demolish part of the northeast corner of the shopping center, and expand the existing retail area by construction of a new building with a larger footprint to make way for the 28,600-square-foot grocery store.
The overall shopping area at Wekiva Riverwalk now covers 229,735 square feet, but that will grow to 259,399 square feet with the retail expansion at the 24.7-acre center, according to filings with SJRWMD.
The shopping center is now anchored by Ross Dress for Less, Marshalls, Bed Bath & Beyond, Petco and United Artists Movie Theaters. Recent improvements to the retail center include new landscaping and color schemes and parking lot. New inline tenants are being brought in.
The new grocery store would add space to the center while also absorbing the space of current tenants Anytime Fitness, Hair Design Institute and Garibaldi's Mexican restaurant.
Residential Property Developments
Developer of Apopka's Oak Pointe project nears groundbreaking after 13 years
Hear from the owner for a lesson in patience, and the opportunity ahead for subcontractors and commercial developers.
Apopka is a growing residential submarket drawing new retailers in recent years. Its population growth between 2010 and 2016 topped 18 percent, outpacing the city of Orlando, Tampa and state of Florida as a whole, according to the U.S. Census Bureau.
Lucky's Market has seven other locations in Greater Orlando further along in the development pipeline, most recently at the Griffin Farm Town Center project in Lake Mary.
In late January, the company announced plans to lease a vacant Sports Authority space near downtown Orlando at 3230 E. Colonial Dr., across from Orlando Fashion Square Mall. A fall opening is estimated there.
The company announced last year it had four new stores in the pipeline for Greater Orlando, not counting the new Apopka site. Those four submarkets named were Hunter's Creek/Kissimmee, Winter Park, Clermont, and the SoDo district in downtown Orlando.
The SoDo location is a mixed-use development at S. Orange and Pineloch avenues where vertical construction is underway. The Winter Park location should be complete late this year in a former Winn-Dixie anchor space in the Unigold Shopping Center, and the Clermont location opened this spring in a former Sweetbay Supermarket space.
Lucky's first area store is in east Orlando at 11750 E. Colonial Dr., on the intersection with Alafaya Trail, which marked its market entry in June 2016.
Have a tip about Central Florida development? Contact me at newsroom@growthspotter.com. Follow GrowthSpotter on Facebook, Twitter and LinkedIn.


Build, craft and explore in Minecraft: PlayStation 3 Edition Discover a world that gives players the freedom to do whatever they want. Dig down into the unknown depths or build incredible structures, from humble homes to grand cityscapes. Survive alone or join forces with friends online and offline. Imagination is the only limit.
First-time players can learn their way around one block at a time in the new tutorial mode.
Supports eight-player online multiplayer and four-player split screen modes.
Includes the free-roaming Creative mode with unlimited supplies to build whatever you can think.


text:
United Airlines In The Dog House (Again) After Puppy Dies On Flight
(ANIMAL NEWS/PET TRAVEL) A 10-month-old French Bulldog puppy died aboard a United Airlines flight Monday after a flight attendant adamantly instructed the unnamed passenger to place the dog in an overhead storage bin.
The airline's policy states all animals traveling inside the cabin should be in a carrier and 'fit completely under the seat in front of the customer and remain there at all times.' Unfortunately in this instance, the carrier was slightly too large to fit under the seat, so the flight attendant advised the pup's parent to store her carrier, with the pet inside, in an overhead bin for the duration of the 3-hour flight from Houston to New York.
But this is hardly the first animal to die aboard a United Airlines flight. In April last year, Simon–destined to be the world's largest rabbit–died while traveling on a United flight between London and Chicago.
In fact, the airline reported more animal deaths on flights last year than any other U.S. carrier. A shocking total of 18 animals died on United flights last year, whereas there were only two animal deaths, respectively, on Delta and American flights in 2017.
There are two important lessons to be learned from this tragedy: 1) pets should never be placed in the overhead bin, and 2) never, ever fly United.
Read on for the details behind this sad story, and click here to check out these 10 tips for safe air travel with pets.– Global Animal
Air travel can be stressful, and even more so when a trip includes an airline flight with the four-legged members of your family. But with a bit of research and preparation, flying with a pet can be safe and comfortable for everyone. Photo Credit: petsugar.com
People Pets, Kelli Bender
Maggie Gremminger is still heartbroken over what she witnessed on March 12 on United Airlines flight #1284 from Houston, Texas, to New York City's Laguardia Airport.
The community relations manager tells PEOPLE that a fellow passenger's dog died on the flight after a flight attendant adamantly instructed the unnamed passenger to store her carrier, with the pet inside, in an overhead storage bin for the duration of the flight. According to Gremminger, 30, the black French bulldog barked from the bin at least 30 minutes into the flight, but was found dead when the owner went to retrieve her dog once the plane had landed.
The tragedy began when the dog owner boarded the plane with a young girl, a baby and the dog in a carrier, which Gremminger says looked like dog carriers that she has seen others use on planes. Gremminger was seated in the row directly behind the party of three and saw the woman place the carrier with the dog inside under the seat in front of her.
I want to help this woman and her daughter. They lost their dog because of an @united flight attendant. My heart is broken. pic.twitter.com/mjXYAhxsAq
— MaggieGremminger (@MaggieGrem) March 13, 2018
Shortly after she was seated, a flight attendant approached the woman with the dog and told her that the bag needed to be moved.
"The flight attendant told the passenger that her bag was blocking part of the aisle. I could not see it, as I was already in my seat, but it sounded like it was somehow not completely fitting beneath the seat in front of her," Gremminger says of the encounter. "After the flight attendant asked her to move it above, the woman adamantly refused, communicating her dog was in the bag. There was some back and forth before finally the flight attendant convinced her to move the carrier to the bin above."
After watching the woman stow her dog, Gremminger says she talked with her seat partner about how "stunned" they were to see a pet be placed in the overhead bin. At this point, Gremminger started to look up information online about whether overhead bins were safe for pets, but was unable to find any information before she had to turn off her phone for takeoff.
I just flew into LGA and witnessed a United flight attendant instruct a passenger to put her dog bag in the overhead bin. It was clearly a dog and while the customer was adamant about leaving it under the seat, the attendant pushed her to do so. (1)
I am disgusted and traumatized. Pets are family. How could a trained flight attendant instruct a passenger to place her dog in that bin. It was her job to understand the plane and it's rules/limitations. (4)
While the situation seemed off to Gremminger, she can understand why the owner ultimately decided to follow the flight attendant's instruction.
"My only thought is that if it had been me, it would have been a hard scenario. The flight attendant is the authority figure, who should be trusted. I was thinking 'maybe there is an improved ventilation system' or something of the sorts," Gremminger says. "Also, the owner had an infant and other daughter. Causing a scene before flight could risk being kicked off the flight. I can only imagine she felt stuck in her decision to comply."
What Gremminger says haunts her the most about this experience was hearing the canine call out from the overhead bin, once while the plane was taking off and again during several moments of turbulence about thirty minutes into the flight. After this the dog was silent.
At the end of the flight the owner opened the overhead bin and placed the dog's carrier on the floor. She soon realized that her pet had died during the flight and was "absolutely shocked and heartbroken."
"A stranger offered to hold her newborn while she sat on the floor, there in the airplane aisle. She was holding her dog and rocking back and forth. Her daughter was also crying," Gremminger says of the scene following the devastating discovery. "People who could not see what was happening were confused as to what was going on. I tried to encourage people to let them off first but it was confusing and so we waited for the rows to filter out. It was absolutely horrible."
The crew, including the flight attendant who instructed the passenger to stow her dog, responded immediately to the woman's distress. Gremminger says the flight attendant who moved the dog seemed "frazzled" and claimed that she was unaware a live animal was in the carrier.
"Many other crew members were contacting additional help and offering a blanket to the young girl who seemed cold. They were confused at how that individual flight attendant could have done this, but did not seem to take any sides or blindly defend," Gremminger adds. "They were professional and did a wonderful job gathering information and being as supportive to the mourning family as possible."
A United Airlines Airbus A319 passenger plane taxis toward a gate at Denver International Airport. Photo Credit: Robert Alexander/Getty Images
PEOPLE reached out to United Airlines for comment on Gremminger's account of the events, and airline spokesperson Maggie Schmerin provided the below statement in response.
"This was a tragic accident that should never have occurred, as pets should never be placed in the overhead bin. We assume full responsibility for this tragedy and express our deepest condolences to the family and are committed to supporting them. We are thoroughly investigating what occurred to prevent this from ever happening again."
According to United's pet policy online, non-service dogs are permitted in the cabin with a service fee of $125 as long as the dog "is in an approved hard-sided or soft-sided kennel. The kennel must fit completely under the seat in front of the customer and remain there at all times."
This is the most recent pet death in a string of animal-related issues for United. Over the past year, the deaths of a famed giant rabbit and several dogs have also been linked to the airline.
More People Pets: http://people.com/pets/dog-dies-on-united-flight-overhead-bin/


Posted by Jay Bartlett May 25, 2017 9:40 pm Categories: Network, New Products. Tags: access, ComNet, Cybersecurity, network, Port Guardian.
ComNet, Communication Networks of Danbury, Connecticut, a USA-based manufacturer of transmission and communication networking equipment, has introduced a new Cybersecurity feature in its latest generation of self-managed switches. The ComNet Port Guardian feature has the capability to physically disconnect a port if unauthorized access is detected.
Posted by Jay Bartlett March 14, 2017 6:06 pm Categories: Corporate News, Cyber Security, IP Cameras. Tags: access, back door, Cybersecurity, Dahua, DVR, firmware update, hack, IP-cameras.
Dahua, the world's second-largest maker of "Internet of Things" devices like security cameras and digital video recorders (DVRs), has shipped a software update that closes a gaping security hole in a wide range of its IP-cameras and DVRs. The vulnerability allows anyone to bypass the login process for these devices and gain remote (and direct) control over vulnerable systems. Adding urgency to the situation, there is now code available online that allows anyone to exploit this bug and commandeer a large number of IoT devices.
Posted by dtaitano June 29, 2014 10:22 pm Categories: Airports / Seaports / Railroads, Biometrics. Tags: access, airport, biometric, morpho, Passenger, Traveler.
Posted by dtaitano June 18, 2014 9:17 pm Categories: Integrated Systems. Tags: access, hd camera, home security, integration, IP-camera, Kronosight, nest, Sentri.


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The transformation of Tacko Fall into a solid NBA center - where does he start?
No one ever knows what to make of the next NBA big man. As the seasons passed, the League welcomed George Mikan, Bill Russell, Wilt Chamberlain, Kareem Abdul-Jabbar - and now Tacko Fall. Mikan was big and talented, but no one ever called him super-athletic. Red Auerbach saw the potential in Russ, and it paid off with 11 Championships for the Celtics. For a player close to 6'10", Big Bill could run, jump, rebound, block shots and defend like no center before him. He was athletic with a Capital A.
Wilt came into the NBA at 7'1" and 250 pounds. His "Wilt the Stilt" nickname fit his physique. He later went up to roughly 300 pounds, and he was super-strong and tough to defend physically. Even Russell needed some occasional help from his back-up, Wayne "The Wall" Embry. Grinding in the paint with Goliath could take a toll. Jabbar brought a new challenge for opposing centers. He was tall, slender and graceful. His sky hook was virtually unstoppable.
All four of these former centers became super stars, and they all brought something new to the table. So does Tacko Fall. The League has never seen anything quite like the Green Giant. At 7'7" and 290 pounds, with a 10"2" reach and 8'2" wingspan, he has already earned Embry's nickname, "The Wall".
Fall the Wall. He stands tall (literally), but I mean he stands quite straight which is good for a person his height. He moves quite well & his turnaround jumper looked like Kevin Garnett's. Tommy Heinsohn would love him. Give him the ball, let him go to work in the paint!
— Brother Andrej (@BriGGySmaLLzb) July 22, 2019
Upon Russell's arrival on the NBA scene, fans were surprised how well he could move for a man of his length. His stride down the court was reminiscent of a gazelle. Fall's run down the court may not quite match Bill's, but he is 10-inches taller and 70-pounds heavier. As the above tweet indicates, he stands straight, runs the court as no one expected and appears to be improving right before our eyes.
Tacko may never reach the heights of Mikan, Russell, Chamberlain or Jabbar, but if he works hard enough, he will find a place in the NBA. His wide shoulders tell me he can get stronger and more muscular, much like Chamberlain. He will need that against the Andre Drummonds of the League. And that strength may help him avoid injury, always a major concern with the super-tall bigs.
The shorter shot-blockers in the NBA need to rely on timing their jumps. Not so important when you have a 10'2" reach. That reach will get Fall a lot of offensive boards and put-backs, but he may need to work on more-aggressive rebounding. This is the NBA after all.
How about 3-point shooting? Forget it - for now. He took no treys in four years of college play. Free throws? A very-poor 43% at UCF. But he can score in close, as his 74% on field goals indicates. His assists/game in college will not rank him with Chris Paul anytime soon (only 0.4 APG).
So there you have it. In a word, Tacko is RAW. But that also means he is a clean slate, and with dedication to the task and hard work, he may never be a super star, but he could have a long NBA career. You can't teach size, and he certainly has that with a good share of grace and fluidity to his movements. Hopefully his strength, distance shooting and passing skills will improve with time. Tacko has become such a fan favorite so quickly, it would be a shame for him not to make it in the NBA.
Follow Tom at @CelticsSentinel, @_Celtics_Center and Facebook
@CelticsSentinel Bill Russell Boston Celtics Kareem Abdul-Jabbar Tacko Fall tomlane UCF Wilt Chamberlain
Labels: @CelticsSentinel Bill Russell Boston Celtics Kareem Abdul-Jabbar Tacko Fall tomlane UCF Wilt Chamberlain
David Anthopoulos July 28, 2019 at 1:26 AM
If Tom has created the "Green Giant" nickname, we should give him credit for thinking up a good one--I rarely look at Twitter, so I'm not really sure whether or not it's been used by others before Tom.
I've seen Tacko's weight listed as high as 311 lbs., but still, he needs to make sure he's not pushed out of the post--I've heard that Danny wants him to put on some more muscle. As should the younger Rob Williams.
And let's put the 43% free throw shooting into perspective. The great defensive center Ben Wallace shot 41.4% for his 16-year NBA career, and only 30% in his rookie season, while Shaq shot 51.1% for his career. Considering that Tacko hasn't been playing the game that long, I suspect that he will be able to pass Shaq with normal growth and development, and he's already ahead of Ben.
Tom Lane July 28, 2019 at 2:06 AM
Believe "Green Giant" is mine. Opened a can of Green Giant asparagus, and the thought hit me. Thanks for mentioning. I had commented in an article back a ways that Green Giant needed to hop on this.
David Anthopoulos July 28, 2019 at 3:44 PM
Though he seems very serious--4 years of college, a Computer Science degree, and 95th percentile in the SATs. So "Jolly" doesn't fit, but "Green" and "Giant" both certainly do.


The Sarasota County School Board has selected four finalists who will be interviewed as the potential successor to Superintendent Lori White, who will retire on Feb. 28, 2017.
Todd Bowden, executive director of career, technical and adult education for the Sarasota County Schools.
Andrew Rynberg, former assistant superintendent of curriculum and instruction with the Indian River Schools (Vero Beach).
Mark Porter, superintendent of the Monroe County Schools (Key West).
Brennan Asplen III, deputy superintendent for academic and student services in St. Johns County (St. Augustine).
A 15-member community advisory committee recommended eight semifinalists to the School Board on Sept. 1, and the board approved that list during its Sept. 6 meeting. By consensus, the board settled on the four finalists following discussions during its Sept. 20 workshop; it voted to approve the list during its regular meeting that afternoon, a news release points out.
The candidates will be invited to be present in Sarasota on Wednesday and Thursday, Oct. 12 and 13, for interviews and conversations with the School Board members, tours of the school district and community, and a public reception, the news release says. The School Board will conduct public interviews on Oct. 12; anyone is welcome to observe those sessions in the School Board Chambers at The Landings, located at 1960 Landings Blvd. in Sarasota, the release adds. The interviews will be streamed live on the district website at SarasotaCountySchools.net and broadcast live on The Education Channel (Comcast 20 and Frontier FiOS 33). On Thursday, Oct. 13, the individual School Board members will interview each candidate privately, the release notes.
A public reception, hosted by the Education Foundation, will begin at 6 p.m. on Oct. 13 in the Riverview High School cafeteria, the release continues. All interested community members are encouraged to attend the event, meet the candidates and ask them questions, the release says. Participants will be invited to write their impressions of the candidates on response cards that will be presented to the School Board members before they make the final selection of the new superintendent, the release adds.
The School Board is expected to appoint the new superintendent during its regular meeting on Oct. 18, which is scheduled for 3 p.m. in the School Board chambers.


At last remortgaging in the UK is rising, but no one is agreed on whether this is a good thing or a bad thing. Certainly, the whole damn mess was partly caused by too many people remortgaging too much too often. It was certainly worsened by the number of people this cycle of remortgaging left buried in untenable negative equity once prices started to fall.
For better or worse some 34,100 borrowers re-mortgaged in August according to data released today by the Council of Mortgage Lenders, a growth of 33% compared to August 2010, and of 9% compared to July. The rise is being put down to a rise in the number of people giving up on the gamble, and fixing at the current low rates.
The CML was quick to point to the fact that last August was one of the worst months on record for remortgaging. It also reminded us -- as if it were necessary -- of the effect fears over just when rates will rise continue to have on remortgaging figures as these fears ebb and flow.
One poignant thing about the data was the stark difference between re-mortgaging growth and that of lending to first time buyers. According to the data just 19,000 first time buyers managed to get a mortgage in August, a growth of 5% compared to last year, and 5% compared to the previous month.
Also of note was the growth in the average loan-to-value ratio from 77% in August 2010 to 80% in August this year. This is hardly likely to stoke the flames of excitement, but according to the CML it is an indication that terms are easing, and 3% of the UK average house price is not far off £5000, depending on which index you read of course.
Paul Smee, director general of the CML, noted that despite the anxieties that turbulence in the eurozone has sparked for lenders, there is clearly demand for mortgage loans in the UK. "With those moving house experiencing a record low in the proportion of their income needed to pay their mortgage interest", he said, "it is clear that the low rate environment is a benefit to those with mortgages, even against the backdrop of the gloom in the wider economy."


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Mission /
The Salvation Army has become aware that comments made on an Australian radio station back in June 2012 have resurfaced and are being reported as being representative of the views of the organisation as a whole. This is incorrect, as we would like to explain.
We would firstly like to make it clear that The Salvation Army does not align itself with comments made in the interview by an officer who is no longer in that spokesperson role. The Salvation Army in Australia promptly apologised for the hurt his comments caused. The Salvation Army in other parts of the world also distanced itself from the comments made. Here are links to a few of the responses – from The Salvation Army in Australia, Canada, the United States and the United Kingdom and Ireland. These were all posted publicly in June 2012, and stand today.
The spokesperson was led into a discussion on Salvation Army Christian beliefs about sin where the interviewers introduced particular texts, including the Romans 1:18-32 passage and some Old Testament Leviticus passages. The interviewers said that The Salvation Army's doctrine book Salvation Story talked about these texts in relation to homosexuality. That is factually untrue. The book discussed was already out of print and superseded by another book, The Handbook of Doctrine by the time of this interview. The newer book also makes no comment on the matter of sexuality - it deals only with the foundational Christian beliefs such as the nature of God, the inspiration of the Bible, the person and work of Jesus Christ, etc.
A diverse range of views on homosexuality exist within The Salvation Army – as among the wider Christian (and non-Christian) community. But no matter where individual Salvationists stand on this matter, The Salvation Army does not permit discrimination on the basis of sexual identity in the delivery of its services or in its employment practices. Our international mission statement is very clear on this point when it says we will 'meet human needs in [Jesus'] name without discrimination'. Anyone who comes through our doors will be welcomed with love and service, based on their need and our capacity to provide.
The Salvation Army stands against homophobia, which victimises people and can reinforce feelings of alienation, loneliness and despair. We want to be an inclusive church community where members of the LGBTQ community find welcome and the encouragement to develop their relationship with God.
We have also been disappointed to notice Photoshopped images that portray Salvation Army fundraisers alongside anti-gay slogans. The alteration of these images maliciously misrepresents people who have freely given their time and energy to raise funds in order that The Salvation Army can provide vital assistance to millions of needy people every year – including many gay, lesbian and transgender people.
While we respect the use of humour and satire, in this instance we are disappointed that the image is presented without making clear that it has been altered.


text:
Listen History
» Artists
» A$ap Twelvyy
A$ap Twelvyy
Jamel Phillips (born May 30, 1989), better known as his stage name ASAP Twelvyy (stylized as A$AP Twelvyy), is an American rapper from Bronx, New York City. He is a member of the hip hop group ASAP Mob, from which he adopted his moniker. In 2014, he released the song "Xscape", the second single to the ASAP Mob's debut album L.O.R.D.. The album was supposed to be released in 2014, but it was officially scrapped. On August 4, 2017, ASAP Twelvyy released his debut album, titled 12. ... (see more)
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Q: draw rectangles and annotate webpage What is the best utility/API to use in order to draw rectangles and annotate a webpage within the browser?
I need the user to be able to draw a box over the existing HTML content and type a message into an annotation for the box.
Similar functionality can be found here - OpenLayers allows you to draw shapes on the map.
EDIT:
I need the solution to be cross-browser, and as lightweight as possible. PrototypeJS is already embedded in this application, if that can be leveraged then that is a bonus.

A: You didn't say what kind of project you are working on, but choose one environment (which supports drawing rectangles over other content) and do everything within that environment. You could, for example, use HTML5 canvas. (If you search through OpenLayers.js a bit, you'll see they did the same)
This is a great resource to get started: http://www.html5canvastutorials.com/

redpajama_set_name: RedPajamaStackExchange


text:
Q: Counting the result LINQ I am using the following
var validLogin = from P in this.DataContext.Persons
                         where P.UserName.Equals(login) && P.Password.Equals(password)
            select new
            {
                P.FirstName,
                P.LastName,
                P.EmailAddress
            };

In this now i want to know, is there any result returned by this query? How to do this.

A: This should work:
if (validLogin.Count() > 0)
{
     //do work
}

A: Don't use Count() - use Any() unless you actually care about the count.
You can actually simplify this a lot, because you don't use the rest of the results either:
bool validLogin = DataContext.Persons.Any(p => p.UserName == login &&
                                               p.Password == password);

The nice thing about Any() is that whatever's processing the query can stop as soon as it's found any matching results - it doesn't need to keep looking for other potential matches. (Count() will work of course, it's just not as efficient, mostly because the operator itself isn't describing what you really care about as accurately.)

A: if (validLogin.Count() > 0){}


blinkx TV allows you to search the web for video and audio clips. Unlike other search providers, blinkx TV not only lets you search using standard keyword and Boolean queries but you can also use conceptual search.
With the greatest respect for the original works of art, the designers of the Paratsone studios in The Netherlands have brought to live famous paintings by lifting images out of the flat surface.


text:
Clay Millen
20 Rams predictions for 2023
By Justin Michael January 11, 2023
DNVR Rams predicts the future for CSU Football.
10 thoughts on CSU football, the 2022 season and the future under Jay Norvell
By Justin Michael November 29, 2022
The 2022 season is officially in the books and although Colorado State finished 3-9 for the second straight year, and missed the postseason for the fifth consecutive year, the feeling in Ram Nation is that the program is trending in the right direction. With that in mind, let's dive right on into it. Here are 10 thoughts on CSU football,…
Air Force beats Colorado State for 13th time in 15 years
Air Force extended its winning streak over Colorado State to six games with a 24-12 victory over the Rams in Colorado Springs Saturday night. The Falcons rushed for 359 yards as a team, the sixth time this season they've gone for at least 300 on the ground. And on a night where Air Force did not complete a single pass…
Colorado State fails to capitalize on early lead, falls to Wyoming in 114th edition of Border War
Pain. For the sixth time in the last seven years, Wyoming will return to Laramie with the Bronze Boot. The Cowboys defeated the Rams 14-13 at Canvas Stadium Saturday night in a game that will sting for the local supporters for quite some time. The Rams had their chances and even led by double digits after jumping out to a…
3 takeaways: CSU shows improvement but comes up short in near upset of San Jose State
By Justin Michael November 6, 2022
Colorado State fell 28-16 to San Jose State in a frustrating game that ultimately got away from the Rams Saturday night. CSU racked up penalties, failed to execute in the Red Zone, and at times struggled to protect the quarterback as the Spartans produced nine sacks and 14 tackles for loss on the night. We already covered all of that…...
Missed chances in the Red Zone come back to bite Colorado State in 28-16 loss at San Jose State
Colorado State could have and probably should have beaten San Jose State on Saturday night. One week after getting bullied by Boise State in a completely one-sided game, CSU looked like a different team this weekend. As 24 point underdogs the Rams went into CEFCU Stadium and gave one of the more experienced and athletic teams in the Mountain West…
Rams takeaways: Colorado State rallies to beat Hawaii
By Justin Michael October 22, 2022
Colorado State found a way on Saturday. The Rams trailed by 10 at halftime and appeared to be on the verge of a disappointing loss to Hawaii. After scoring just 3 points in the first 30 minutes of action though, CSU outscored the visiting Warriors 14-0 to come away with a 17-13 victory at Canvas Stadium. Going back to 1965,…
DNVR Rams Podcast: Why Clay Millen's first month was encouraging compared to former quarterbacks at Nevada & a look at how former Rams are faring with their new programs
Justin Michael dives into Clay Millen's first month as a starting quarterback and compares it to Carson Strong (2019) and Ty Gangi (2017) at Nevada. Justin talks about how their situations are similar, as well as how they're different. He also breaks down some goals for Millen in the second half of the year based on how those under guys…
Rams takeaways: Early missed opportunities come back to bite Colorado State in narrow loss to Utah State
Colorado State's 17-13 loss to Utah State was frustrating on multiple levels. The Rams had a great opportunity to move to 2-0 in conference play with another winnable game coming up against Hawaii. After losing the first four games of the year, the Rams had a chance to put themselves in position to at least try and make a run…...
Jay Norvell gives update on QB situation for Colorado State — how Clay Millen can benefit from missed time
Clay Millen is still dealing with a shoulder injury and is expected to miss another week. During his weekly press conference on Monday afternoon, head coach Jay Norvell said that the hope is for the Rams to have their starting quarterback back for the Hawaii game. Unfortunately though, it's looking like the sprained AC Joint that Millen suffered against Sacramento…...
An update on Colorado State's quarterback situation
By Justin Michael September 28, 2022
Colorado State's highly anticipated trip to Nevada is just around the corner and at this point it's unclear who will be the team's starting quarterback for the conference opener. Clay Millen has been the team's No. 1 QB since coming over to CSU with the staff last winter and his numbers have been fairly solid for a first-time starter. In…...
DNVR Rams Podcast: Matt Mumme on Clay Millen's progress, Around the Mountain West, & Justin's TV debut
Justin Michael starts the show by talking about how excited he is to be the sideline reporter for the TV broadcast of Saturday's game against Sacramento State. He then previews his interview with associate HC/QB coach Matt Mumme. Finally, Justin gives an update on the other teams around the Mountain West. DON'T FORGET TO SUBSCRIBE! iTunes Link | Stitcher Link |Spotify Link | Google Play…
Matt Mumme opens up on Clay Millen's progression, battling adversity, and the long journey ahead
Clay Millen is three starts into his collegiate career and while the Rams have not managed to win a game yet, the redshirt freshman quarterback is showing growth. Millen is completing 71% of his pass attempts (61-of-85), which is about 8% more than Carson Strong at this stage back in 2019, an encouraging sign according to associate head coach Matt…...
3 takeaways: Another slow start dooms CSU in a 38-7 loss at Washington State
Colorado State fell to 0-3 for the first time since 2010 with a 38-7 loss at Washington State (3-0) Saturday. Dating back to last season, CSU's losing streak of nine games is currently the longest in the country. Starting with the Rams' first offensive series of the game, a three-and-out that ended with WAZZU linebacker Daiyon Henley sacking Clay Millen…...
Trust the process — it's going to take time to rebuild CSU Football the right way
Colorado State has a long road ahead. Saturday's home opener was a reminder that while the program is on the right track under Jay Norvell, the results are probably going to be a mixed bag in the first season. It's easy to get ahead of ourselves, get caught up in the excitement of a new era, and all that comes…...
DNVR Rams Podcast: Takeaways from a humbling loss to Middle Tennessee
Justin Michael dives into what went wrong for Colorado State in a 34-19 loss to Middle Tennessee. Justin discusses the state of the offensive line, an inconsistent passing game so far and what to keep an eye on moving forward. He also explains why despite a frustrating outcome, Ram fans must remain patient with a staff that is doing the…
Colorado State struggles to find consistency in a disappointing performance against Middle Tennessee
Saturday's home opener was a familiar scene for Ram Nation. Colorado State (0-2) was favored at home, the students showed up ready to rock, and by halftime most of them were ready to move on with their day. Considering the Rams were held scoreless in the first half for the second straight week and at one point trailed 34-0 — but…...
DNVR Rams Podcast: Week 1 takeaways from CSU-Michigan and the entire Mountain West
By Justin Michael September 5, 2022
Justin Michael breaks down the Week 1 matchup with Michigan and explains what the Rams did well, as well as where they need to improve. Justin gives his thoughts on the gap between the top couple of schools in the country and everyone else. Finally, he ends the show with some thoughts on the Mountain West slate, including a wonky…
3 takeaways: Colorado State's offense shows progress in live reps
By Justin Michael August 19, 2022
Colorado State held its second of three scrimmages on Wednesday and although head coach Jay Norvell could not attend due to a positive COVID-19 test, the Rams, led by offensive coordinator Matt Mumme and defensive coordinator Freddie Banks, were able to get in live reps on a gorgeous morning in Fort Collins. "You hate not having your fearless leader," Mumme…...
DNVR Rams Podcast: Roddy to workout for the Raptors, Mountain West QB power rankings & more!
By Justin Michael May 31, 2022
Justin Michael gives an update on David Roddy, before talking about what he sees his role in the NBA being, and what elements of his game he could improve upon with one more year at CSU. Justin then gives his preseason power rankings for the Mountain West quarterbacks going into 2022. DON'T FORGET TO SUBSCRIBE! iTunes Link | Stitcher Link |Spotify Link | Google Play
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